The following is a list of integrals (antiderivative functions ) of trigonometric functions . For antiderivatives involving both exponential and trigonometric functions, see List of integrals of exponential functions . For a complete list of antiderivative functions, see Lists of integrals . For the special antiderivatives involving trigonometric functions, see Trigonometric integral .[1] [2]
Generally, if the function
sin
x
{\displaystyle \sin x}
cos
x
{\displaystyle \cos x}
∫
a
cos
n
x
d
x
=
a
n
sin
n
x
+
C
{\displaystyle \int a\cos nx\,dx={\frac {a}{n}}\sin nx+C}
In all formulas the constant a is assumed to be nonzero, and C denotes the constant of integration .
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Evaluating Integrals With Trigonometric Functions
Indefinite Integral - Basic Integration Rules, Problems, Formulas, Trig Functions, Calculus
#50. INTEGRAL OF TRIGONOMETRIC FUNCTIONS | CALCULUS | INTEGRATION
Calculus 2 Lecture 7.2: Techniques For Trigonometric Integrals
Integrands involving only sine
∫
sin
a
x
d
x
=
−
1
a
cos
a
x
+
C
{\displaystyle \int \sin ax\,dx=-{\frac {1}{a}}\cos ax+C}
∫
sin
2
a
x
d
x
=
x
2
−
1
4
a
sin
2
a
x
+
C
=
x
2
−
1
2
a
sin
a
x
cos
a
x
+
C
{\displaystyle \int \sin ^{2}{ax}\,dx={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}-{\frac {1}{2a}}\sin ax\cos ax+C}
∫
sin
3
a
x
d
x
=
cos
3
a
x
12
a
−
3
cos
a
x
4
a
+
C
{\displaystyle \int \sin ^{3}{ax}\,dx={\frac {\cos 3ax}{12a}}-{\frac {3\cos ax}{4a}}+C}
∫
x
sin
2
a
x
d
x
=
x
2
4
−
x
4
a
sin
2
a
x
−
1
8
a
2
cos
2
a
x
+
C
{\displaystyle \int x\sin ^{2}{ax}\,dx={\frac {x^{2}}{4}}-{\frac {x}{4a}}\sin 2ax-{\frac {1}{8a^{2}}}\cos 2ax+C}
∫
x
2
sin
2
a
x
d
x
=
x
3
6
−
(
x
2
4
a
−
1
8
a
3
)
sin
2
a
x
−
x
4
a
2
cos
2
a
x
+
C
{\displaystyle \int x^{2}\sin ^{2}{ax}\,dx={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C}
∫
x
sin
a
x
d
x
=
sin
a
x
a
2
−
x
cos
a
x
a
+
C
{\displaystyle \int x\sin ax\,dx={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C}
∫
(
sin
b
1
x
)
(
sin
b
2
x
)
d
x
=
sin
(
(
b
2
−
b
1
)
x
)
2
(
b
2
−
b
1
)
−
sin
(
(
b
1
+
b
2
)
x
)
2
(
b
1
+
b
2
)
+
C
(for
|
b
1
|
≠
|
b
2
|
)
{\displaystyle \int (\sin b_{1}x)(\sin b_{2}x)\,dx={\frac {\sin((b_{2}-b_{1})x)}{2(b_{2}-b_{1})}}-{\frac {\sin((b_{1}+b_{2})x)}{2(b_{1}+b_{2})}}+C\qquad {\mbox{(for }}|b_{1}|\neq |b_{2}|{\mbox{)}}}
∫
sin
n
a
x
d
x
=
−
sin
n
−
1
a
x
cos
a
x
n
a
+
n
−
1
n
∫
sin
n
−
2
a
x
d
x
(for
n
>
0
)
{\displaystyle \int \sin ^{n}{ax}\,dx=-{\frac {\sin ^{n-1}ax\cos ax}{na}}+{\frac {n-1}{n}}\int \sin ^{n-2}ax\,dx\qquad {\mbox{(for }}n>0{\mbox{)}}}
∫
d
x
sin
a
x
=
−
1
a
ln
|
csc
a
x
+
cot
a
x
|
+
C
{\displaystyle \int {\frac {dx}{\sin ax}}=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C}
∫
d
x
sin
n
a
x
=
cos
a
x
a
(
1
−
n
)
sin
n
−
1
a
x
+
n
−
2
n
−
1
∫
d
x
sin
n
−
2
a
x
(for
n
>
1
)
{\displaystyle \int {\frac {dx}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}}
∫
x
n
sin
a
x
d
x
=
−
x
n
a
cos
a
x
+
n
a
∫
x
n
−
1
cos
a
x
d
x
=
∑
k
=
0
2
k
≤
n
(
−
1
)
k
+
1
x
n
−
2
k
a
1
+
2
k
n
!
(
n
−
2
k
)
!
cos
a
x
+
∑
k
=
0
2
k
+
1
≤
n
(
−
1
)
k
x
n
−
1
−
2
k
a
2
+
2
k
n
!
(
n
−
2
k
−
1
)
!
sin
a
x
=
−
∑
k
=
0
n
x
n
−
k
a
1
+
k
n
!
(
n
−
k
)
!
cos
(
a
x
+
k
π
2
)
(for
n
>
0
)
{\displaystyle {\begin{aligned}\int x^{n}\sin ax\,dx&=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\,dx\\&=\sum _{k=0}^{2k\leq n}(-1)^{k+1}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\cos ax+\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-1-2k}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\sin ax\\&=-\sum _{k=0}^{n}{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\cos \left(ax+k{\frac {\pi }{2}}\right)\qquad {\mbox{(for }}n>0{\mbox{)}}\end{aligned}}}
∫
sin
a
x
x
d
x
=
∑
n
=
0
∞
(
−
1
)
n
(
a
x
)
2
n
+
1
(
2
n
+
1
)
⋅
(
2
n
+
1
)
!
+
C
{\displaystyle \int {\frac {\sin ax}{x}}\,dx=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!}}+C}
∫
sin
a
x
x
n
d
x
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
+
a
n
−
1
∫
cos
a
x
x
n
−
1
d
x
{\displaystyle \int {\frac {\sin ax}{x^{n}}}\,dx=-{\frac {\sin ax}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}\int {\frac {\cos ax}{x^{n-1}}}\,dx}
∫
sin
(
a
x
2
+
b
x
+
c
)
d
x
=
{
a
π
2
cos
(
b
2
−
4
a
c
4
a
)
S
(
2
a
x
+
b
2
a
π
)
+
a
π
2
sin
(
b
2
−
4
a
c
4
a
)
C
(
2
a
x
+
b
2
a
π
)
t
o
b
2
−
4
a
c
>
0
a
π
2
cos
(
b
2
−
4
a
c
4
a
)
S
(
2
a
x
+
b
2
a
π
)
−
a
π
2
sin
(
b
2
−
4
a
c
4
a
)
C
(
2
a
x
+
b
2
a
π
)
t
o
b
2
−
4
a
c
<
0
f
o
r
a
╱
=
0
,
a
>
0
{\displaystyle \int {\sin {\mathrm {(} }{ax}^{2}\mathrm {+} {bx}\mathrm {+} {c}{\mathrm {)} }{dx}}\mathrm {=} \left\{{\begin{aligned}&{{\sqrt {a}}{\sqrt {\frac {\mathit {\pi }}{2}}}\cos \left({\frac {{b}^{2}\mathrm {-} {4}{ac}}{4a}}\right){S}\left({\frac {{2}{ax}\mathrm {+} {b}}{\sqrt {{2}{a}{\mathit {\pi }}}}}\right)\mathrm {+} {\sqrt {a}}{\sqrt {\frac {\mathit {\pi }}{2}}}\sin \left({\frac {{b}^{2}\mathrm {-} {4}{ac}}{4a}}\right){C}\left({\frac {{2}{ax}\mathrm {+} {b}}{\sqrt {{2}{a}{\mathit {\pi }}}}}\right)\;{to}\;{b}^{2}\mathrm {-} {4}{ac}\;{\mathrm {>} }\;{0}}\\&{{\sqrt {a}}{\sqrt {\frac {\mathit {\pi }}{2}}}\cos \left({\frac {{b}^{2}\mathrm {-} {4}{ac}}{4a}}\right){S}\left({\frac {{2}{ax}\mathrm {+} {b}}{\sqrt {{2}{a}{\mathit {\pi }}}}}\right)\mathrm {-} {\sqrt {a}}{\sqrt {\frac {\mathit {\pi }}{2}}}\sin \left({\frac {{b}^{2}\mathrm {-} {4}{ac}}{4a}}\right){C}\left({\frac {{2}{ax}\mathrm {+} {b}}{\sqrt {{2}{a}{\mathit {\pi }}}}}\right)\;{to}\;{b}^{2}\mathrm {-} {4}{ac}\;{\mathrm {<} }\;{0}}\end{aligned}}\right.\;\;{for}\;{a}\diagup \!\!\!\!{\mathrm {=} }{0}{\mathrm {,} }\;{a}{\mathrm {>} }{0}}
∫
d
x
1
±
sin
a
x
=
1
a
tan
(
a
x
2
∓
π
4
)
+
C
{\displaystyle \int {\frac {dx}{1\pm \sin ax}}={\frac {1}{a}}\tan \left({\frac {ax}{2}}\mp {\frac {\pi }{4}}\right)+C}
∫
x
d
x
1
+
sin
a
x
=
x
a
tan
(
a
x
2
−
π
4
)
+
2
a
2
ln
|
cos
(
a
x
2
−
π
4
)
|
+
C
{\displaystyle \int {\frac {x\,dx}{1+\sin ax}}={\frac {x}{a}}\tan \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{a^{2}}}\ln \left|\cos \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)\right|+C}
∫
x
d
x
1
−
sin
a
x
=
x
a
cot
(
π
4
−
a
x
2
)
+
2
a
2
ln
|
sin
(
π
4
−
a
x
2
)
|
+
C
{\displaystyle \int {\frac {x\,dx}{1-\sin ax}}={\frac {x}{a}}\cot \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)+{\frac {2}{a^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)\right|+C}
∫
sin
a
x
d
x
1
±
sin
a
x
=
±
x
+
1
a
tan
(
π
4
∓
a
x
2
)
+
C
{\displaystyle \int {\frac {\sin ax\,dx}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C}
Integrands involving only cosine
∫
cos
a
x
d
x
=
1
a
sin
a
x
+
C
{\displaystyle \int \cos ax\,dx={\frac {1}{a}}\sin ax+C}
∫
cos
2
a
x
d
x
=
x
2
+
1
4
a
sin
2
a
x
+
C
=
x
2
+
1
2
a
sin
a
x
cos
a
x
+
C
{\displaystyle \int \cos ^{2}{ax}\,dx={\frac {x}{2}}+{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}+{\frac {1}{2a}}\sin ax\cos ax+C}
∫
cos
n
a
x
d
x
=
cos
n
−
1
a
x
sin
a
x
n
a
+
n
−
1
n
∫
cos
n
−
2
a
x
d
x
(for
n
>
0
)
{\displaystyle \int \cos ^{n}ax\,dx={\frac {\cos ^{n-1}ax\sin ax}{na}}+{\frac {n-1}{n}}\int \cos ^{n-2}ax\,dx\qquad {\mbox{(for }}n>0{\mbox{)}}}
∫
x
cos
a
x
d
x
=
cos
a
x
a
2
+
x
sin
a
x
a
+
C
{\displaystyle \int x\cos ax\,dx={\frac {\cos ax}{a^{2}}}+{\frac {x\sin ax}{a}}+C}
∫
x
2
cos
2
a
x
d
x
=
x
3
6
+
(
x
2
4
a
−
1
8
a
3
)
sin
2
a
x
+
x
4
a
2
cos
2
a
x
+
C
{\displaystyle \int x^{2}\cos ^{2}{ax}\,dx={\frac {x^{3}}{6}}+\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax+{\frac {x}{4a^{2}}}\cos 2ax+C}
∫
x
n
cos
a
x
d
x
=
x
n
sin
a
x
a
−
n
a
∫
x
n
−
1
sin
a
x
d
x
=
∑
k
=
0
2
k
+
1
≤
n
(
−
1
)
k
x
n
−
2
k
−
1
a
2
+
2
k
n
!
(
n
−
2
k
−
1
)
!
cos
a
x
+
∑
k
=
0
2
k
≤
n
(
−
1
)
k
x
n
−
2
k
a
1
+
2
k
n
!
(
n
−
2
k
)
!
sin
a
x
=
∑
k
=
0
n
(
−
1
)
⌊
k
/
2
⌋
x
n
−
k
a
1
+
k
n
!
(
n
−
k
)
!
cos
(
a
x
−
(
−
1
)
k
+
1
2
π
2
)
=
∑
k
=
0
n
x
n
−
k
a
1
+
k
n
!
(
n
−
k
)
!
sin
(
a
x
+
k
π
2
)
(for
n
>
0
)
{\displaystyle {\begin{aligned}\int x^{n}\cos ax\,dx&={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\,dx\\&=\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-2k-1}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\cos ax+\sum _{k=0}^{2k\leq n}(-1)^{k}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\sin ax\\&=\sum _{k=0}^{n}(-1)^{\lfloor k/2\rfloor }{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\cos \left(ax-{\frac {(-1)^{k}+1}{2}}{\frac {\pi }{2}}\right)\\&=\sum _{k=0}^{n}{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\sin \left(ax+k{\frac {\pi }{2}}\right)\qquad {\mbox{(for }}n>0{\mbox{)}}\end{aligned}}}
∫
cos
a
x
x
d
x
=
ln
|
a
x
|
+
∑
k
=
1
∞
(
−
1
)
k
(
a
x
)
2
k
2
k
⋅
(
2
k
)
!
+
C
{\displaystyle \int {\frac {\cos ax}{x}}\,dx=\ln |ax|+\sum _{k=1}^{\infty }(-1)^{k}{\frac {(ax)^{2k}}{2k\cdot (2k)!}}+C}
∫
cos
a
x
x
n
d
x
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
∫
sin
a
x
x
n
−
1
d
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos ax}{x^{n}}}\,dx=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}\,dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
d
x
cos
a
x
=
1
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
{\displaystyle \int {\frac {dx}{\cos ax}}={\frac {1}{a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}
∫
d
x
cos
n
a
x
=
sin
a
x
a
(
n
−
1
)
cos
n
−
1
a
x
+
n
−
2
n
−
1
∫
d
x
cos
n
−
2
a
x
(for
n
>
1
)
{\displaystyle \int {\frac {dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}}
∫
d
x
1
+
cos
a
x
=
1
a
tan
a
x
2
+
C
{\displaystyle \int {\frac {dx}{1+\cos ax}}={\frac {1}{a}}\tan {\frac {ax}{2}}+C}
∫
d
x
1
−
cos
a
x
=
−
1
a
cot
a
x
2
+
C
{\displaystyle \int {\frac {dx}{1-\cos ax}}=-{\frac {1}{a}}\cot {\frac {ax}{2}}+C}
∫
x
d
x
1
+
cos
a
x
=
x
a
tan
a
x
2
+
2
a
2
ln
|
cos
a
x
2
|
+
C
{\displaystyle \int {\frac {x\,dx}{1+\cos ax}}={\frac {x}{a}}\tan {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\cos {\frac {ax}{2}}\right|+C}
∫
x
d
x
1
−
cos
a
x
=
−
x
a
cot
a
x
2
+
2
a
2
ln
|
sin
a
x
2
|
+
C
{\displaystyle \int {\frac {x\,dx}{1-\cos ax}}=-{\frac {x}{a}}\cot {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\sin {\frac {ax}{2}}\right|+C}
∫
cos
a
x
d
x
1
+
cos
a
x
=
x
−
1
a
tan
a
x
2
+
C
{\displaystyle \int {\frac {\cos ax\,dx}{1+\cos ax}}=x-{\frac {1}{a}}\tan {\frac {ax}{2}}+C}
∫
cos
a
x
d
x
1
−
cos
a
x
=
−
x
−
1
a
cot
a
x
2
+
C
{\displaystyle \int {\frac {\cos ax\,dx}{1-\cos ax}}=-x-{\frac {1}{a}}\cot {\frac {ax}{2}}+C}
∫
(
cos
a
1
x
)
(
cos
a
2
x
)
d
x
=
sin
(
(
a
2
−
a
1
)
x
)
2
(
a
2
−
a
1
)
+
sin
(
(
a
2
+
a
1
)
x
)
2
(
a
2
+
a
1
)
+
C
(for
|
a
1
|
≠
|
a
2
|
)
{\displaystyle \int (\cos a_{1}x)(\cos a_{2}x)\,dx={\frac {\sin((a_{2}-a_{1})x)}{2(a_{2}-a_{1})}}+{\frac {\sin((a_{2}+a_{1})x)}{2(a_{2}+a_{1})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}}
Integrands involving only tangent
∫
tan
a
x
d
x
=
−
1
a
ln
|
cos
a
x
|
+
C
=
1
a
ln
|
sec
a
x
|
+
C
{\displaystyle \int \tan ax\,dx=-{\frac {1}{a}}\ln |\cos ax|+C={\frac {1}{a}}\ln |\sec ax|+C}
∫
tan
2
x
d
x
=
tan
x
−
x
+
C
{\displaystyle \int \tan ^{2}{x}\,dx=\tan {x}-x+C}
∫
tan
n
a
x
d
x
=
1
a
(
n
−
1
)
tan
n
−
1
a
x
−
∫
tan
n
−
2
a
x
d
x
(for
n
≠
1
)
{\displaystyle \int \tan ^{n}ax\,dx={\frac {1}{a(n-1)}}\tan ^{n-1}ax-\int \tan ^{n-2}ax\,dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
d
x
q
tan
a
x
+
p
=
1
p
2
+
q
2
(
p
x
+
q
a
ln
|
q
sin
a
x
+
p
cos
a
x
|
)
+
C
(for
p
2
+
q
2
≠
0
)
{\displaystyle \int {\frac {dx}{q\tan ax+p}}={\frac {1}{p^{2}+q^{2}}}(px+{\frac {q}{a}}\ln |q\sin ax+p\cos ax|)+C\qquad {\mbox{(for }}p^{2}+q^{2}\neq 0{\mbox{)}}}
∫
d
x
tan
a
x
±
1
=
±
x
2
+
1
2
a
ln
|
sin
a
x
±
cos
a
x
|
+
C
{\displaystyle \int {\frac {dx}{\tan ax\pm 1}}=\pm {\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax\pm \cos ax|+C}
∫
tan
a
x
d
x
tan
a
x
±
1
=
x
2
∓
1
2
a
ln
|
sin
a
x
±
cos
a
x
|
+
C
{\displaystyle \int {\frac {\tan ax\,dx}{\tan ax\pm 1}}={\frac {x}{2}}\mp {\frac {1}{2a}}\ln |\sin ax\pm \cos ax|+C}
Integrands involving only secant
∫
sec
a
x
d
x
=
1
a
ln
|
sec
a
x
+
tan
a
x
|
+
C
=
1
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
=
1
a
artanh
(
sin
a
x
)
+
C
{\displaystyle \int \sec {ax}\,dx={\frac {1}{a}}\ln {\left|\sec {ax}+\tan {ax}\right|}+C={\frac {1}{a}}\ln {\left|\tan {\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)}\right|}+C={\frac {1}{a}}\operatorname {artanh} {\left(\sin {ax}\right)}+C}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}{x}\,dx=\tan {x}+C}
∫
sec
3
x
d
x
=
1
2
sec
x
tan
x
+
1
2
ln
|
sec
x
+
tan
x
|
+
C
.
{\displaystyle \int \sec ^{3}{x}\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+C.}
∫
sec
n
a
x
d
x
=
sec
n
−
2
a
x
tan
a
x
a
(
n
−
1
)
+
n
−
2
n
−
1
∫
sec
n
−
2
a
x
d
x
(for
n
≠
1
)
{\displaystyle \int \sec ^{n}{ax}\,dx={\frac {\sec ^{n-2}{ax}\tan {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{ax}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}}
∫
d
x
sec
x
+
1
=
x
−
tan
x
2
+
C
{\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\tan {\frac {x}{2}}+C}
∫
d
x
sec
x
−
1
=
−
x
−
cot
x
2
+
C
{\displaystyle \int {\frac {dx}{\sec {x}-1}}=-x-\cot {\frac {x}{2}}+C}
∫
sin
x
cos
x
=
∫
tan
x
{\displaystyle \int {\frac {\sin {x}}{\cos {x}}}=\int \tan {x}}
Integrands involving only cosecant
∫
csc
a
x
d
x
=
−
1
a
ln
|
csc
a
x
+
cot
a
x
|
+
C
=
1
a
ln
|
csc
a
x
−
cot
a
x
|
+
C
=
1
a
ln
|
tan
(
a
x
2
)
|
+
C
{\displaystyle \int \csc {ax}\,dx=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C={\frac {1}{a}}\ln {\left|\csc {ax}-\cot {ax}\right|}+C={\frac {1}{a}}\ln {\left|\tan {\left({\frac {ax}{2}}\right)}\right|}+C}
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int \csc ^{2}{x}\,dx=-\cot {x}+C}
∫
csc
3
x
d
x
=
−
1
2
csc
x
cot
x
−
1
2
ln
|
csc
x
+
cot
x
|
+
C
=
−
1
2
csc
x
cot
x
+
1
2
ln
|
csc
x
−
cot
x
|
+
C
{\displaystyle \int \csc ^{3}{x}\,dx=-{\frac {1}{2}}\csc x\cot x-{\frac {1}{2}}\ln |\csc x+\cot x|+C=-{\frac {1}{2}}\csc x\cot x+{\frac {1}{2}}\ln |\csc x-\cot x|+C}
∫
csc
n
a
x
d
x
=
−
csc
n
−
2
a
x
cot
a
x
a
(
n
−
1
)
+
n
−
2
n
−
1
∫
csc
n
−
2
a
x
d
x
(for
n
≠
1
)
{\displaystyle \int \csc ^{n}{ax}\,dx=-{\frac {\csc ^{n-2}{ax}\cot {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{ax}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}}
∫
d
x
csc
x
+
1
=
x
−
2
cot
x
2
+
1
+
C
{\displaystyle \int {\frac {dx}{\csc {x}+1}}=x-{\frac {2}{\cot {\frac {x}{2}}+1}}+C}
∫
d
x
csc
x
−
1
=
−
x
+
2
cot
x
2
−
1
+
C
{\displaystyle \int {\frac {dx}{\csc {x}-1}}=-x+{\frac {2}{\cot {\frac {x}{2}}-1}}+C}
Integrands involving only cotangent
∫
cot
a
x
d
x
=
1
a
ln
|
sin
a
x
|
+
C
{\displaystyle \int \cot ax\,dx={\frac {1}{a}}\ln |\sin ax|+C}
∫
cot
2
x
d
x
=
−
cot
x
−
x
+
C
{\displaystyle \int \cot ^{2}{x}\,dx=-\cot {x}-x+C}
∫
cot
n
a
x
d
x
=
−
1
a
(
n
−
1
)
cot
n
−
1
a
x
−
∫
cot
n
−
2
a
x
d
x
(for
n
≠
1
)
{\displaystyle \int \cot ^{n}ax\,dx=-{\frac {1}{a(n-1)}}\cot ^{n-1}ax-\int \cot ^{n-2}ax\,dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
d
x
1
+
cot
a
x
=
∫
tan
a
x
d
x
tan
a
x
+
1
=
x
2
−
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
{\displaystyle \int {\frac {dx}{1+\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax+1}}={\frac {x}{2}}-{\frac {1}{2a}}\ln |\sin ax+\cos ax|+C}
∫
d
x
1
−
cot
a
x
=
∫
tan
a
x
d
x
tan
a
x
−
1
=
x
2
+
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
{\displaystyle \int {\frac {dx}{1-\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax-1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax-\cos ax|+C}
Integrands involving both sine and cosine An integral that is a rational function of the sine and cosine can be evaluated using Bioche's rules .
∫
d
x
cos
a
x
±
sin
a
x
=
1
a
2
ln
|
tan
(
a
x
2
±
π
8
)
|
+
C
{\displaystyle \int {\frac {dx}{\cos ax\pm \sin ax}}={\frac {1}{a{\sqrt {2}}}}\ln \left|\tan \left({\frac {ax}{2}}\pm {\frac {\pi }{8}}\right)\right|+C}
∫
d
x
(
cos
a
x
±
sin
a
x
)
2
=
1
2
a
tan
(
a
x
∓
π
4
)
+
C
{\displaystyle \int {\frac {dx}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C}
∫
d
x
(
cos
x
+
sin
x
)
n
=
1
2
(
n
−
1
)
(
sin
x
−
cos
x
(
cos
x
+
sin
x
)
n
−
1
+
(
n
−
2
)
∫
d
x
(
cos
x
+
sin
x
)
n
−
2
)
{\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{2(n-1)}}\left({\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}+(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right)}
∫
cos
a
x
d
x
cos
a
x
+
sin
a
x
=
x
2
+
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
{\displaystyle \int {\frac {\cos ax\,dx}{\cos ax+\sin ax}}={\frac {x}{2}}+{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C}
∫
cos
a
x
d
x
cos
a
x
−
sin
a
x
=
x
2
−
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
{\displaystyle \int {\frac {\cos ax\,dx}{\cos ax-\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C}
∫
sin
a
x
d
x
cos
a
x
+
sin
a
x
=
x
2
−
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
{\displaystyle \int {\frac {\sin ax\,dx}{\cos ax+\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C}
∫
sin
a
x
d
x
cos
a
x
−
sin
a
x
=
−
x
2
−
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
{\displaystyle \int {\frac {\sin ax\,dx}{\cos ax-\sin ax}}=-{\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C}
∫
cos
a
x
d
x
(
sin
a
x
)
(
1
+
cos
a
x
)
=
−
1
4
a
tan
2
a
x
2
+
1
2
a
ln
|
tan
a
x
2
|
+
C
{\displaystyle \int {\frac {\cos ax\,dx}{(\sin ax)(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}
∫
cos
a
x
d
x
(
sin
a
x
)
(
1
−
cos
a
x
)
=
−
1
4
a
cot
2
a
x
2
−
1
2
a
ln
|
tan
a
x
2
|
+
C
{\displaystyle \int {\frac {\cos ax\,dx}{(\sin ax)(1-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}
∫
sin
a
x
d
x
(
cos
a
x
)
(
1
+
sin
a
x
)
=
1
4
a
cot
2
(
a
x
2
+
π
4
)
+
1
2
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
{\displaystyle \int {\frac {\sin ax\,dx}{(\cos ax)(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}
∫
sin
a
x
d
x
(
cos
a
x
)
(
1
−
sin
a
x
)
=
1
4
a
tan
2
(
a
x
2
+
π
4
)
−
1
2
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
{\displaystyle \int {\frac {\sin ax\,dx}{(\cos ax)(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}
∫
(
sin
a
x
)
(
cos
a
x
)
d
x
=
1
2
a
sin
2
a
x
+
C
{\displaystyle \int (\sin ax)(\cos ax)\,dx={\frac {1}{2a}}\sin ^{2}ax+C}
∫
(
sin
a
1
x
)
(
cos
a
2
x
)
d
x
=
−
cos
(
(
a
1
−
a
2
)
x
)
2
(
a
1
−
a
2
)
−
cos
(
(
a
1
+
a
2
)
x
)
2
(
a
1
+
a
2
)
+
C
(for
|
a
1
|
≠
|
a
2
|
)
{\displaystyle \int (\sin a_{1}x)(\cos a_{2}x)\,dx=-{\frac {\cos((a_{1}-a_{2})x)}{2(a_{1}-a_{2})}}-{\frac {\cos((a_{1}+a_{2})x)}{2(a_{1}+a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}}
∫
(
sin
n
a
x
)
(
cos
a
x
)
d
x
=
1
a
(
n
+
1
)
sin
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int (\sin ^{n}ax)(\cos ax)\,dx={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}}
∫
(
sin
a
x
)
(
cos
n
a
x
)
d
x
=
−
1
a
(
n
+
1
)
cos
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int (\sin ax)(\cos ^{n}ax)\,dx=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}}
∫
(
sin
n
a
x
)
(
cos
m
a
x
)
d
x
=
−
(
sin
n
−
1
a
x
)
(
cos
m
+
1
a
x
)
a
(
n
+
m
)
+
n
−
1
n
+
m
∫
(
sin
n
−
2
a
x
)
(
cos
m
a
x
)
d
x
(for
m
,
n
>
0
)
=
(
sin
n
+
1
a
x
)
(
cos
m
−
1
a
x
)
a
(
n
+
m
)
+
m
−
1
n
+
m
∫
(
sin
n
a
x
)
(
cos
m
−
2
a
x
)
d
x
(for
m
,
n
>
0
)
{\displaystyle {\begin{aligned}\int (\sin ^{n}ax)(\cos ^{m}ax)\,dx&=-{\frac {(\sin ^{n-1}ax)(\cos ^{m+1}ax)}{a(n+m)}}+{\frac {n-1}{n+m}}\int (\sin ^{n-2}ax)(\cos ^{m}ax)\,dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\\&={\frac {(\sin ^{n+1}ax)(\cos ^{m-1}ax)}{a(n+m)}}+{\frac {m-1}{n+m}}\int (\sin ^{n}ax)(\cos ^{m-2}ax)\,dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\end{aligned}}}
∫
d
x
(
sin
a
x
)
(
cos
a
x
)
=
1
a
ln
|
tan
a
x
|
+
C
{\displaystyle \int {\frac {dx}{(\sin ax)(\cos ax)}}={\frac {1}{a}}\ln \left|\tan ax\right|+C}
∫
d
x
(
sin
a
x
)
(
cos
n
a
x
)
=
1
a
(
n
−
1
)
cos
n
−
1
a
x
+
∫
d
x
(
sin
a
x
)
(
cos
n
−
2
a
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {dx}{(\sin ax)(\cos ^{n}ax)}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {dx}{(\sin ax)(\cos ^{n-2}ax)}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
d
x
(
sin
n
a
x
)
(
cos
a
x
)
=
−
1
a
(
n
−
1
)
sin
n
−
1
a
x
+
∫
d
x
(
sin
n
−
2
a
x
)
(
cos
a
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {dx}{(\sin ^{n}ax)(\cos ax)}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {dx}{(\sin ^{n-2}ax)(\cos ax)}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
sin
a
x
d
x
cos
n
a
x
=
1
a
(
n
−
1
)
cos
n
−
1
a
x
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ax\,dx}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
sin
2
a
x
d
x
cos
a
x
=
−
1
a
sin
a
x
+
1
a
ln
|
tan
(
π
4
+
a
x
2
)
|
+
C
{\displaystyle \int {\frac {\sin ^{2}ax\,dx}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {ax}{2}}\right)\right|+C}
∫
sin
2
a
x
d
x
cos
n
a
x
=
sin
a
x
a
(
n
−
1
)
cos
n
−
1
a
x
−
1
n
−
1
∫
d
x
cos
n
−
2
a
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{2}ax\,dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
sin
2
x
1
+
cos
2
x
d
x
=
2
arctangant
(
tan
x
2
)
−
x
(for x in
]
−
π
2
;
+
π
2
[
)
=
2
arctangant
(
tan
x
2
)
−
arctangant
(
tan
x
)
(this time x being any real number
)
{\displaystyle {\begin{aligned}\int {\frac {\sin ^{2}x}{1+\cos ^{2}x}}\,dx&={\sqrt {2}}\operatorname {arctangant} \left({\frac {\tan x}{\sqrt {2}}}\right)-x\qquad {\mbox{(for x in}}]-{\frac {\pi }{2}};+{\frac {\pi }{2}}[{\mbox{)}}\\&={\sqrt {2}}\operatorname {arctangant} \left({\frac {\tan x}{\sqrt {2}}}\right)-\operatorname {arctangant} \left(\tan x\right)\qquad {\mbox{(this time x being any real number }}{\mbox{)}}\end{aligned}}}
∫
sin
n
a
x
d
x
cos
a
x
=
−
sin
n
−
1
a
x
a
(
n
−
1
)
+
∫
sin
n
−
2
a
x
d
x
cos
a
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}ax\,dx}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\,dx}{\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
sin
n
a
x
d
x
cos
m
a
x
=
{
sin
n
+
1
a
x
a
(
m
−
1
)
cos
m
−
1
a
x
−
n
−
m
+
2
m
−
1
∫
sin
n
a
x
d
x
cos
m
−
2
a
x
(for
m
≠
1
)
sin
n
−
1
a
x
a
(
m
−
1
)
cos
m
−
1
a
x
−
n
−
1
m
−
1
∫
sin
n
−
2
a
x
d
x
cos
m
−
2
a
x
(for
m
≠
1
)
−
sin
n
−
1
a
x
a
(
n
−
m
)
cos
m
−
1
a
x
+
n
−
1
n
−
m
∫
sin
n
−
2
a
x
d
x
cos
m
a
x
(for
m
≠
n
)
{\displaystyle \int {\frac {\sin ^{n}ax\,dx}{\cos ^{m}ax}}={\begin{cases}{\frac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}ax\,dx}{\cos ^{m-2}ax}}&{\mbox{(for }}m\neq 1{\mbox{)}}\\{\frac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}ax\,dx}{\cos ^{m-2}ax}}&{\mbox{(for }}m\neq 1{\mbox{)}}\\-{\frac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}ax\,dx}{\cos ^{m}ax}}&{\mbox{(for }}m\neq n{\mbox{)}}\end{cases}}}
∫
cos
a
x
d
x
sin
n
a
x
=
−
1
a
(
n
−
1
)
sin
n
−
1
a
x
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos ax\,dx}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
cos
2
a
x
d
x
sin
a
x
=
1
a
(
cos
a
x
+
ln
|
tan
a
x
2
|
)
+
C
{\displaystyle \int {\frac {\cos ^{2}ax\,dx}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left|\tan {\frac {ax}{2}}\right|\right)+C}
∫
cos
2
a
x
d
x
sin
n
a
x
=
−
1
n
−
1
(
cos
a
x
a
sin
n
−
1
a
x
+
∫
d
x
sin
n
−
2
a
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos ^{2}ax\,dx}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax}}+\int {\frac {dx}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
cos
n
a
x
d
x
sin
m
a
x
=
{
−
cos
n
+
1
a
x
a
(
m
−
1
)
sin
m
−
1
a
x
−
n
−
m
+
2
m
−
1
∫
cos
n
a
x
d
x
sin
m
−
2
a
x
(for
n
≠
1
)
−
cos
n
−
1
a
x
a
(
m
−
1
)
sin
m
−
1
a
x
−
n
−
1
m
−
1
∫
cos
n
−
2
a
x
d
x
sin
m
−
2
a
x
(for
m
≠
1
)
cos
n
−
1
a
x
a
(
n
−
m
)
sin
m
−
1
a
x
+
n
−
1
n
−
m
∫
cos
n
−
2
a
x
d
x
sin
m
a
x
(for
m
≠
n
)
{\displaystyle \int {\frac {\cos ^{n}ax\,dx}{\sin ^{m}ax}}={\begin{cases}-{\frac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\cos ^{n}ax\,dx}{\sin ^{m-2}ax}}&{\mbox{(for }}n\neq 1{\mbox{)}}\\-{\frac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\cos ^{n-2}ax\,dx}{\sin ^{m-2}ax}}&{\mbox{(for }}m\neq 1{\mbox{)}}\\{\frac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\cos ^{n-2}ax\,dx}{\sin ^{m}ax}}&{\mbox{(for }}m\neq n{\mbox{)}}\end{cases}}}
Integrands involving both sine and tangent
∫
(
sin
a
x
)
(
tan
a
x
)
d
x
=
1
a
(
ln
|
sec
a
x
+
tan
a
x
|
−
sin
a
x
)
+
C
{\displaystyle \int (\sin ax)(\tan ax)\,dx={\frac {1}{a}}(\ln |\sec ax+\tan ax|-\sin ax)+C}
∫
tan
n
a
x
d
x
sin
2
a
x
=
1
a
(
n
−
1
)
tan
n
−
1
(
a
x
)
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\tan ^{n}ax\,dx}{\sin ^{2}ax}}={\frac {1}{a(n-1)}}\tan ^{n-1}(ax)+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
Integrand involving both cosine and tangent
∫
tan
n
a
x
d
x
cos
2
a
x
=
1
a
(
n
+
1
)
tan
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int {\frac {\tan ^{n}ax\,dx}{\cos ^{2}ax}}={\frac {1}{a(n+1)}}\tan ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}}
Integrand involving both sine and cotangent
∫
cot
n
a
x
d
x
sin
2
a
x
=
−
1
a
(
n
+
1
)
cot
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int {\frac {\cot ^{n}ax\,dx}{\sin ^{2}ax}}=-{\frac {1}{a(n+1)}}\cot ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}}
∫
cot
n
a
x
d
x
cos
2
a
x
=
1
a
(
1
−
n
)
tan
1
−
n
a
x
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\cot ^{n}ax\,dx}{\cos ^{2}ax}}={\frac {1}{a(1-n)}}\tan ^{1-n}ax+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
Integrand involving both secant and tangent
∫
(
sec
x
)
(
tan
x
)
d
x
=
sec
x
+
C
{\displaystyle \int (\sec x)(\tan x)\,dx=\sec x+C}
∫
(
csc
x
)
(
cot
x
)
d
x
=
−
csc
x
+
C
{\displaystyle \int (\csc x)(\cot x)\,dx=-\csc x+C}
Integrals in a quarter period Using the beta function
B
(
a
,
b
)
{\displaystyle B(a,b)}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
1
2
B
(
n
+
1
2
,
1
2
)
=
{
n
−
1
n
⋅
n
−
3
n
−
2
⋯
3
4
⋅
1
2
⋅
π
2
,
if
n
is even
n
−
1
n
⋅
n
−
3
n
−
2
⋯
4
5
⋅
2
3
,
if
n
is odd and more than 1
1
,
if
n
=
1
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx={\frac {1}{2}}B\left({\frac {n+1}{2}},{\frac {1}{2}}\right)={\begin{cases}{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot {\frac {\pi }{2}},&{\text{if }}n{\text{ is even}}\\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {4}{5}}\cdot {\frac {2}{3}},&{\text{if }}n{\text{ is odd and more than 1}}\\1,&{\text{if }}n=1\end{cases}}}
Integrals with symmetric limits
∫
−
c
c
sin
x
d
x
=
0
{\displaystyle \int _{-c}^{c}\sin {x}\,dx=0}
∫
−
c
c
cos
x
d
x
=
2
∫
0
c
cos
x
d
x
=
2
∫
−
c
0
cos
x
d
x
=
2
sin
c
{\displaystyle \int _{-c}^{c}\cos {x}\,dx=2\int _{0}^{c}\cos {x}\,dx=2\int _{-c}^{0}\cos {x}\,dx=2\sin {c}}
∫
−
c
c
tan
x
d
x
=
0
{\displaystyle \int _{-c}^{c}\tan {x}\,dx=0}
∫
−
a
2
a
2
x
2
cos
2
n
π
x
a
d
x
=
a
3
(
n
2
π
2
−
6
)
24
n
2
π
2
(for
n
=
1
,
3
,
5...
)
{\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\,dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=1,3,5...{\mbox{)}}}
∫
−
a
2
a
2
x
2
sin
2
n
π
x
a
d
x
=
a
3
(
n
2
π
2
−
6
(
−
1
)
n
)
24
n
2
π
2
=
a
3
24
(
1
−
6
(
−
1
)
n
n
2
π
2
)
(for
n
=
1
,
2
,
3
,
.
.
.
)
{\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\,dx={\frac {a^{3}(n^{2}\pi ^{2}-6(-1)^{n})}{24n^{2}\pi ^{2}}}={\frac {a^{3}}{24}}(1-6{\frac {(-1)^{n}}{n^{2}\pi ^{2}}})\qquad {\mbox{(for }}n=1,2,3,...{\mbox{)}}}
Integral over a full circle
∫
0
2
π
sin
2
m
+
1
x
cos
n
x
d
x
=
0
n
,
m
∈
Z
{\displaystyle \int _{0}^{2\pi }\sin ^{2m+1}{x}\cos ^{n}{x}\,dx=0\!\qquad n,m\in \mathbb {Z} }
∫
0
2
π
sin
m
x
cos
2
n
+
1
x
d
x
=
0
n
,
m
∈
Z
{\displaystyle \int _{0}^{2\pi }\sin ^{m}{x}\cos ^{2n+1}{x}\,dx=0\!\qquad n,m\in \mathbb {Z} }
See also References
^ ^ Bresock, Krista, "Student Understanding of the Definite Integral When Solving Calculus Volume Problems" (2022). Graduate Theses, Dissertations, and Problem Reports. 11491. https://researchrepository.wvu.edu/etd/11491
This page was last edited on 12 April 2024, at 16:42
![{\displaystyle {\begin{aligned}\int {\frac {\sin ^{2}x}{1+\cos ^{2}x}}\,dx&={\sqrt {2}}\operatorname {arctangant} \left({\frac {\tan x}{\sqrt {2}}}\right)-x\qquad {\mbox{(for x in}}]-{\frac {\pi }{2}};+{\frac {\pi }{2}}[{\mbox{)}}\\&={\sqrt {2}}\operatorname {arctangant} \left({\frac {\tan x}{\sqrt {2}}}\right)-\operatorname {arctangant} \left(\tan x\right)\qquad {\mbox{(this time x being any real number }}{\mbox{)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99bc35b310db277a8b20f736913c8178097758b6)
This page was last edited on 12 April 2024, at 16:42