redaktə
∫
sin
(
a
x
+
b
)
d
x
=
−
1
a
cos
(
a
x
+
b
)
+
C
{\displaystyle \int \sin(ax+b)\,dx=-{\frac {1}{a}}\cos(ax+b)+C}
∫
cos
(
a
x
+
b
)
d
x
=
1
a
sin
(
a
x
+
b
)
+
C
{\displaystyle \int \cos(ax+b)\,dx={\frac {1}{a}}\sin(ax+b)+C}
∫
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}
∫
cotan
(
a
x
)
d
x
=
1
a
ln
|
sin
(
a
x
)
|
+
C
{\displaystyle \int \operatorname {cotan} (ax)\,dx={\frac {1}{a}}\ln |\sin(ax)|+C}
∫
sin
(
x
)
d
x
=
−
cos
(
x
)
+
C
{\displaystyle \int \sin(x)\,dx=-\cos(x)+C}
∫
cos
(
x
)
d
x
=
sin
(
x
)
+
C
{\displaystyle \int \cos(x)\,dx=\sin(x)+C}
∫
tan
(
x
)
d
x
=
−
ln
|
cos
(
x
)
|
+
C
=
ln
|
sec
(
x
)
|
+
C
{\displaystyle \int \tan(x)\,dx=-\ln |\cos(x)|+C=\ln |\sec(x)|+C}
∫
cotan
(
x
)
d
x
=
ln
|
sin
(
x
)
|
+
C
=
−
ln
|
cosec
(
x
)
|
+
C
{\displaystyle \int \operatorname {cotan} (x)\,dx=\ln |\sin(x)|+C=-\ln |\operatorname {cosec} (x)|+C}
∫
sin
c
x
d
x
=
−
1
c
cos
c
x
{\displaystyle \int \sin cx\;dx=-{\frac {1}{c}}\cos cx\,\!}
∫
sin
n
c
x
d
x
=
−
sin
n
−
1
c
x
cos
c
x
n
c
+
n
−
1
n
∫
sin
n
−
2
c
x
d
x
(
n
>
0
)
{\displaystyle \int \sin ^{n}cx\;dx=-{\frac {\sin ^{n-1}cx\cos cx}{nc}}+{\frac {n-1}{n}}\int \sin ^{n-2}cx\;dx\qquad {\mbox{( }}n>0{\mbox{)}}\,\!}
∫
x
sin
c
x
d
x
=
sin
c
x
c
2
−
x
cos
c
x
c
{\displaystyle \int x\sin cx\;dx={\frac {\sin cx}{c^{2}}}-{\frac {x\cos cx}{c}}\,\!}
∫
x
2
sin
c
x
d
x
=
2
cos
c
x
c
3
+
2
x
sin
c
x
c
2
−
x
2
cos
c
x
c
{\displaystyle \int x^{2}\sin cx\;dx={\frac {2\cos cx}{c^{3}}}+{\frac {2x\sin cx}{c^{2}}}-{\frac {x^{2}\cos cx}{c}}\,\!}
∫
x
3
sin
c
x
d
x
=
−
6
sin
c
x
c
4
+
6
x
cos
c
x
c
3
+
3
x
2
sin
c
x
c
2
−
x
3
cos
c
x
c
{\displaystyle \int x^{3}\sin cx\;dx=-{\frac {6\sin cx}{c^{4}}}+{\frac {6x\cos cx}{c^{3}}}+{\frac {3x^{2}\sin cx}{c^{2}}}-{\frac {x^{3}\cos cx}{c}}\,\!}
∫
x
4
sin
c
x
d
x
=
−
24
cos
c
x
c
5
−
24
x
sin
c
x
c
4
+
12
x
2
cos
c
x
c
3
+
4
x
3
sin
c
x
c
2
−
x
4
cos
c
x
c
{\displaystyle \int x^{4}\sin cx\;dx=-{\frac {24\cos cx}{c^{5}}}-{\frac {24x\sin cx}{c^{4}}}+{\frac {12x^{2}\cos cx}{c^{3}}}+{\frac {4x^{3}\sin cx}{c^{2}}}-{\frac {x^{4}\cos cx}{c}}\,\!}
∫
x
5
sin
c
x
d
x
=
120
sin
c
x
c
6
−
120
x
cos
c
x
c
5
−
60
x
2
sin
c
x
c
4
+
20
x
3
cos
c
x
c
3
+
5
x
4
sin
c
x
c
2
−
x
5
cos
c
x
c
{\displaystyle \int x^{5}\sin cx\;dx={\frac {120\sin cx}{c^{6}}}-{\frac {120x\cos cx}{c^{5}}}-{\frac {60x^{2}\sin cx}{c^{4}}}+{\frac {20x^{3}\cos cx}{c^{3}}}+{\frac {5x^{4}\sin cx}{c^{2}}}-{\frac {x^{5}\cos cx}{c}}\,\!}
∫
x
n
sin
c
x
d
x
=
n
!
⋅
sin
c
x
[
x
n
−
1
c
2
⋅
(
n
−
1
)
!
−
x
n
−
3
c
4
⋅
(
n
−
3
)
!
+
x
n
−
5
c
6
⋅
(
n
−
5
)
!
−
.
.
.
]
−
−
n
!
⋅
cos
c
x
[
x
n
c
⋅
n
!
−
x
n
−
2
c
3
⋅
(
n
−
2
)
!
+
x
n
−
4
c
5
⋅
(
n
−
4
)
!
−
.
.
.
]
{\displaystyle {\begin{aligned}\int x^{n}\sin cx\;dx&=n!\cdot \sin cx\left[{\frac {x^{n-1}}{c^{2}\cdot (n-1)!}}-{\frac {x^{n-3}}{c^{4}\cdot (n-3)!}}+{\frac {x^{n-5}}{c^{6}\cdot (n-5)!}}-...\right]-\\&-n!\cdot \cos cx\left[{\frac {x^{n}}{c\cdot n!}}-{\frac {x^{n-2}}{c^{3}\cdot (n-2)!}}+{\frac {x^{n-4}}{c^{5}\cdot (n-4)!}}-...\right]\end{aligned}}}
∫
x
n
sin
c
x
d
x
=
−
x
n
c
cos
c
x
+
n
c
∫
x
n
−
1
cos
c
x
d
x
(
n
>
0
)
{\displaystyle \int x^{n}\sin cx\;dx=-{\frac {x^{n}}{c}}\cos cx+{\frac {n}{c}}\int x^{n-1}\cos cx\;dx\qquad {\mbox{( }}n>0{\mbox{)}}\,\!}
∫
sin
c
x
x
d
x
=
∑
i
=
0
∞
(
−
1
)
i
(
c
x
)
2
i
+
1
(
2
i
+
1
)
⋅
(
2
i
+
1
)
!
{\displaystyle \int {\frac {\sin cx}{x}}dx=\sum _{i=0}^{\infty }(-1)^{i}{\frac {(cx)^{2i+1}}{(2i+1)\cdot (2i+1)!}}\,\!}
∫
sin
c
x
x
n
d
x
=
−
sin
c
x
(
n
−
1
)
x
n
−
1
+
c
n
−
1
∫
cos
c
x
x
n
−
1
d
x
{\displaystyle \int {\frac {\sin cx}{x^{n}}}dx=-{\frac {\sin cx}{(n-1)x^{n-1}}}+{\frac {c}{n-1}}\int {\frac {\cos cx}{x^{n-1}}}dx\,\!}
∫
d
x
sin
c
x
=
1
c
ln
|
tg
c
x
2
|
{\displaystyle \int {\frac {dx}{\sin cx}}={\frac {1}{c}}\ln \left|\operatorname {tg} {\frac {cx}{2}}\right|}
∫
d
x
sin
n
c
x
=
cos
c
x
c
(
1
−
n
)
sin
n
−
1
c
x
+
n
−
2
n
−
1
∫
d
x
sin
n
−
2
c
x
(
n
>
1
)
{\displaystyle \int {\frac {dx}{\sin ^{n}cx}}={\frac {\cos cx}{c(1-n)\sin ^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}cx}}\qquad {\mbox{( }}n>1{\mbox{)}}\,\!}
∫
d
x
1
±
sin
c
x
=
1
c
tg
(
c
x
2
∓
π
4
)
{\displaystyle \int {\frac {dx}{1\pm \sin cx}}={\frac {1}{c}}\operatorname {tg} \left({\frac {cx}{2}}\mp {\frac {\pi }{4}}\right)}
∫
x
d
x
1
+
sin
c
x
=
x
c
tg
(
c
x
2
−
π
4
)
+
2
c
2
ln
|
cos
(
c
x
2
−
π
4
)
|
{\displaystyle \int {\frac {x\;dx}{1+\sin cx}}={\frac {x}{c}}\operatorname {tg} \left({\frac {cx}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{c^{2}}}\ln \left|\cos \left({\frac {cx}{2}}-{\frac {\pi }{4}}\right)\right|}
∫
x
d
x
1
−
sin
c
x
=
x
c
ctg
(
π
4
−
c
x
2
)
+
2
c
2
ln
|
sin
(
π
4
−
c
x
2
)
|
{\displaystyle \int {\frac {x\;dx}{1-\sin cx}}={\frac {x}{c}}\operatorname {ctg} \left({\frac {\pi }{4}}-{\frac {cx}{2}}\right)+{\frac {2}{c^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {cx}{2}}\right)\right|}
∫
sin
c
x
d
x
1
±
sin
c
x
=
±
x
+
1
c
tg
(
π
4
∓
c
x
2
)
{\displaystyle \int {\frac {\sin cx\;dx}{1\pm \sin cx}}=\pm x+{\frac {1}{c}}\operatorname {tg} \left({\frac {\pi }{4}}\mp {\frac {cx}{2}}\right)}
∫
sin
c
1
x
sin
c
2
x
d
x
=
sin
(
c
1
−
c
2
)
x
2
(
c
1
−
c
2
)
−
sin
(
c
1
+
c
2
)
x
2
(
c
1
+
c
2
)
(
|
c
1
|
≠
|
c
2
|
)
{\displaystyle \int \sin c_{1}x\sin c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}-{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad {\mbox{( }}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}
∫
cos
c
x
d
x
=
1
c
sin
c
x
{\displaystyle \int \cos cx\;dx={\frac {1}{c}}\sin cx\,\!}
∫
cos
n
c
x
d
x
=
cos
n
−
1
c
x
sin
c
x
n
c
+
n
−
1
n
∫
cos
n
−
2
c
x
d
x
(
n
>
0
)
{\displaystyle \int \cos ^{n}cx\;dx={\frac {\cos ^{n-1}cx\sin cx}{nc}}+{\frac {n-1}{n}}\int \cos ^{n-2}cx\;dx\qquad {\mbox{( }}n>0{\mbox{)}}\,\!}
∫
x
cos
c
x
d
x
=
cos
c
x
c
2
+
x
sin
c
x
c
{\displaystyle \int x\cos cx\;dx={\frac {\cos cx}{c^{2}}}+{\frac {x\sin cx}{c}}\,\!}
∫
x
n
cos
c
x
d
x
=
x
n
sin
c
x
c
−
n
c
∫
x
n
−
1
sin
c
x
d
x
{\displaystyle \int x^{n}\cos cx\;dx={\frac {x^{n}\sin cx}{c}}-{\frac {n}{c}}\int x^{n-1}\sin cx\;dx\,\!}
∫
cos
c
x
x
d
x
=
ln
|
c
x
|
+
∑
i
=
1
∞
(
−
1
)
i
(
c
x
)
2
i
2
i
⋅
(
2
i
)
!
{\displaystyle \int {\frac {\cos cx}{x}}dx=\ln |cx|+\sum _{i=1}^{\infty }(-1)^{i}{\frac {(cx)^{2i}}{2i\cdot (2i)!}}\,\!}
∫
cos
c
x
x
n
d
x
=
−
cos
c
x
(
n
−
1
)
x
n
−
1
−
c
n
−
1
∫
sin
c
x
x
n
−
1
d
x
(
n
≠
1
)
{\displaystyle \int {\frac {\cos cx}{x^{n}}}dx=-{\frac {\cos cx}{(n-1)x^{n-1}}}-{\frac {c}{n-1}}\int {\frac {\sin cx}{x^{n-1}}}dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
cos
c
x
=
1
c
ln
|
tg
(
c
x
2
+
π
4
)
|
{\displaystyle \int {\frac {dx}{\cos cx}}={\frac {1}{c}}\ln \left|\operatorname {tg} \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}
∫
d
x
cos
n
c
x
=
sin
c
x
c
(
n
−
1
)
c
o
s
n
−
1
c
x
+
n
−
2
n
−
1
∫
d
x
cos
n
−
2
c
x
(
n
>
1
)
{\displaystyle \int {\frac {dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)cos^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{( }}n>1{\mbox{)}}\,\!}
∫
d
x
1
+
cos
c
x
=
1
c
tg
c
x
2
{\displaystyle \int {\frac {dx}{1+\cos cx}}={\frac {1}{c}}\operatorname {tg} {\frac {cx}{2}}\,\!}
∫
d
x
1
−
cos
c
x
=
−
1
c
ctg
c
x
2
{\displaystyle \int {\frac {dx}{1-\cos cx}}=-{\frac {1}{c}}\operatorname {ctg} {\frac {cx}{2}}\,\!}
∫
x
d
x
1
+
cos
c
x
=
x
c
tg
c
x
2
+
2
c
2
ln
|
cos
c
x
2
|
{\displaystyle \int {\frac {x\;dx}{1+\cos cx}}={\frac {x}{c}}\operatorname {tg} {cx}{2}+{\frac {2}{c^{2}}}\ln \left|\cos {\frac {cx}{2}}\right|}
∫
x
d
x
1
−
cos
c
x
=
−
x
x
ctg
c
x
2
+
2
c
2
ln
|
sin
c
x
2
|
{\displaystyle \int {\frac {x\;dx}{1-\cos cx}}=-{\frac {x}{x}}\operatorname {ctg} {cx}{2}+{\frac {2}{c^{2}}}\ln \left|\sin {\frac {cx}{2}}\right|}
∫
cos
c
x
d
x
1
+
cos
c
x
=
x
−
1
c
tg
c
x
2
{\displaystyle \int {\frac {\cos cx\;dx}{1+\cos cx}}=x-{\frac {1}{c}}\operatorname {tg} {\frac {cx}{2}}\,\!}
∫
cos
c
x
d
x
1
−
cos
c
x
=
−
x
−
1
c
ctg
c
x
2
{\displaystyle \int {\frac {\cos cx\;dx}{1-\cos cx}}=-x-{\frac {1}{c}}\operatorname {ctg} {\frac {cx}{2}}\,\!}
∫
cos
c
1
x
cos
c
2
x
d
x
=
sin
(
c
1
−
c
2
)
x
2
(
c
1
−
c
2
)
+
sin
(
c
1
+
c
2
)
x
2
(
c
1
+
c
2
)
(
|
c
1
|
≠
|
c
2
|
)
{\displaystyle \int \cos c_{1}x\cos c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}+{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad {\mbox{( }}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}
∫
tg
c
x
d
x
=
−
1
c
ln
|
cos
c
x
|
{\displaystyle \int \operatorname {tg} cx\;dx=-{\frac {1}{c}}\ln |\cos cx|\,\!}
∫
tg
n
c
x
d
x
=
1
c
(
n
−
1
)
tg
n
−
1
c
x
−
∫
tg
n
−
2
c
x
d
x
(
n
≠
1
)
{\displaystyle \int \operatorname {tg} ^{n}cx\;dx={\frac {1}{c(n-1)}}\operatorname {tg} ^{n-1}cx-\int \operatorname {tg} ^{n-2}cx\;dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
tg
c
x
+
1
=
x
2
+
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {dx}{\operatorname {tg} cx+1}}={\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx+\cos cx|\,\!}
∫
d
x
tg
c
x
−
1
=
−
x
2
+
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {dx}{\operatorname {tg} cx-1}}=-{\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx-\cos cx|\,\!}
∫
tg
c
x
d
x
tg
c
x
+
1
=
x
2
−
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {\operatorname {tg} cx\;dx}{\operatorname {tg} cx+1}}={\frac {x}{2}}-{\frac {1}{2c}}\ln |\sin cx+\cos cx|\,\!}
∫
tg
c
x
d
x
tg
c
x
−
1
=
x
2
+
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {\operatorname {tg} cx\;dx}{\operatorname {tg} cx-1}}={\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx-\cos cx|\,\!}
∫
ctg
c
x
d
x
=
1
c
ln
|
sin
c
x
|
{\displaystyle \int \operatorname {ctg} cx\;dx={\frac {1}{c}}\ln |\sin cx|\,\!}
∫
ctg
n
c
x
d
x
=
−
1
c
(
n
−
1
)
ctg
n
−
1
c
x
−
∫
ctg
n
−
2
c
x
d
x
(
n
≠
1
)
{\displaystyle \int \operatorname {ctg} ^{n}cx\;dx=-{\frac {1}{c(n-1)}}\operatorname {ctg} ^{n-1}cx-\int \operatorname {ctg} ^{n-2}cx\;dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
1
+
ctg
c
x
=
∫
tg
c
x
d
x
tg
c
x
+
1
{\displaystyle \int {\frac {dx}{1+\operatorname {ctg} cx}}=\int {\frac {\operatorname {tg} cx\;dx}{\operatorname {tg} cx+1}}\,\!}
∫
d
x
1
−
ctg
c
x
=
∫
tg
c
x
d
x
tg
c
x
−
1
{\displaystyle \int {\frac {dx}{1-\operatorname {ctg} cx}}=\int {\frac {\operatorname {tg} cx\;dx}{\operatorname {tg} cx-1}}\,\!}
∫
sec
c
x
d
x
=
1
c
ln
|
sec
c
x
+
tg
c
x
|
{\displaystyle \int \sec {cx}\,dx={\frac {1}{c}}\ln {\left|\sec {cx}+\operatorname {tg} {cx}\right|}}
∫
sec
n
c
x
d
x
=
sec
n
−
1
c
x
sin
c
x
c
(
n
−
1
)
+
n
−
2
n
−
1
∫
sec
n
−
2
c
x
d
x
(
n
≠
1
)
{\displaystyle \int \sec ^{n}{cx}\,dx={\frac {\sec ^{n-1}{cx}\sin {cx}}{c(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{cx}\,dx\qquad {\mbox{ ( }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
sec
x
+
1
=
x
−
tg
x
2
{\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\operatorname {tg} {\frac {x}{2}}}
∫
csc
c
x
d
x
=
−
1
c
ln
|
csc
c
x
+
ctg
c
x
|
{\displaystyle \int \csc {cx}\,dx=-{\frac {1}{c}}\ln {\left|\csc {cx}+\operatorname {ctg} {cx}\right|}}
∫
csc
n
c
x
d
x
=
−
csc
n
−
1
c
x
cos
c
x
c
(
n
−
1
)
+
n
−
2
n
−
1
∫
csc
n
−
2
c
x
d
x
(
n
≠
1
)
{\displaystyle \int \csc ^{n}{cx}\,dx=-{\frac {\csc ^{n-1}{cx}\cos {cx}}{c(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{cx}\,dx\qquad {\mbox{ ( }}n\neq 1{\mbox{)}}\,\!}
redaktə
∫
d
x
cos
c
x
±
sin
c
x
=
1
c
2
ln
|
tg
(
c
x
2
±
π
8
)
|
{\displaystyle \int {\frac {dx}{\cos cx\pm \sin cx}}={\frac {1}{c{\sqrt {2}}}}\ln \left|\operatorname {tg} \left({\frac {cx}{2}}\pm {\frac {\pi }{8}}\right)\right|}
∫
d
x
(
cos
c
x
±
sin
c
x
)
2
=
1
2
c
tg
(
c
x
∓
π
4
)
{\displaystyle \int {\frac {dx}{(\cos cx\pm \sin cx)^{2}}}={\frac {1}{2c}}\operatorname {tg} \left(cx\mp {\frac {\pi }{4}}\right)}
∫
d
x
(
cos
x
+
sin
x
)
n
=
1
n
−
1
(
sin
x
−
cos
x
(
cos
x
+
sin
x
)
n
−
1
−
2
(
n
−
2
)
∫
d
x
(
cos
x
+
sin
x
)
n
−
2
)
{\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left({\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right)}
∫
cos
c
x
d
x
cos
c
x
+
sin
c
x
=
x
2
+
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {\cos cx\;dx}{\cos cx+\sin cx}}={\frac {x}{2}}+{\frac {1}{2c}}\ln \left|\sin cx+\cos cx\right|}
∫
cos
c
x
d
x
cos
c
x
−
sin
c
x
=
x
2
−
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {\cos cx\;dx}{\cos cx-\sin cx}}={\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx-\cos cx\right|}
∫
sin
c
x
d
x
cos
c
x
+
sin
c
x
=
x
2
−
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx+\sin cx}}={\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx+\cos cx\right|}
∫
sin
c
x
d
x
cos
c
x
−
sin
c
x
=
−
x
2
−
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx-\sin cx}}=-{\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx-\cos cx\right|}
∫
cos
c
x
d
x
sin
c
x
(
1
+
cos
c
x
)
=
−
1
4
c
tg
2
c
x
2
+
1
2
c
ln
|
tg
c
x
2
|
{\displaystyle \int {\frac {\cos cx\;dx}{\sin cx(1+\cos cx)}}=-{\frac {1}{4c}}\operatorname {tg} ^{2}{\frac {cx}{2}}+{\frac {1}{2c}}\ln \left|\operatorname {tg} {\frac {cx}{2}}\right|}
∫
cos
c
x
d
x
sin
c
x
(
1
+
−
cos
c
x
)
=
−
1
4
c
ctg
2
c
x
2
−
1
2
c
ln
|
tg
c
x
2
|
{\displaystyle \int {\frac {\cos cx\;dx}{\sin cx(1+-\cos cx)}}=-{\frac {1}{4c}}\operatorname {ctg} ^{2}{\frac {cx}{2}}-{\frac {1}{2c}}\ln \left|\operatorname {tg} {\frac {cx}{2}}\right|}
∫
sin
c
x
d
x
cos
c
x
(
1
+
sin
c
x
)
=
1
4
c
ctg
2
(
c
x
2
+
π
4
)
+
1
2
c
ln
|
tg
(
c
x
2
+
π
4
)
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx(1+\sin cx)}}={\frac {1}{4c}}\operatorname {ctg} ^{2}\left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2c}}\ln \left|\operatorname {tg} \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}
∫
sin
c
x
d
x
cos
c
x
(
1
−
sin
c
x
)
=
1
4
c
tg
2
(
c
x
2
+
π
4
)
−
1
2
c
ln
|
tg
(
c
x
2
+
π
4
)
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx(1-\sin cx)}}={\frac {1}{4c}}\operatorname {tg} ^{2}\left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2c}}\ln \left|\operatorname {tg} \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}
∫
sin
c
x
cos
c
x
d
x
=
1
2
c
sin
2
c
x
{\displaystyle \int \sin cx\cos cx\;dx={\frac {1}{2c}}\sin ^{2}cx\,\!}
∫
sin
c
1
x
cos
c
2
x
d
x
=
−
cos
(
c
1
+
c
2
)
x
2
(
c
1
+
c
2
)
−
cos
(
c
1
−
c
2
)
x
2
(
c
1
−
c
2
)
(
|
c
1
|
≠
|
c
2
|
)
{\displaystyle \int \sin c_{1}x\cos c_{2}x\;dx=-{\frac {\cos(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}-{\frac {\cos(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}\qquad {\mbox{( }}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}
∫
sin
n
c
x
cos
c
x
d
x
=
1
c
(
n
+
1
)
sin
n
+
1
c
x
(
n
≠
1
)
{\displaystyle \int \sin ^{n}cx\cos cx\;dx={\frac {1}{c(n+1)}}\sin ^{n+1}cx\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
sin
c
x
cos
n
c
x
d
x
=
−
1
c
(
n
+
1
)
cos
n
+
1
c
x
(
n
≠
1
)
{\displaystyle \int \sin cx\cos ^{n}cx\;dx=-{\frac {1}{c(n+1)}}\cos ^{n+1}cx\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
cos
m
c
x
d
x
=
−
sin
n
−
1
c
x
cos
m
+
1
c
x
c
(
n
+
m
)
+
n
−
1
n
+
m
∫
sin
n
−
2
c
x
cos
m
c
x
d
x
(
m
,
n
>
0
)
{\displaystyle \int \sin ^{n}cx\cos ^{m}cx\;dx=-{\frac {\sin ^{n-1}cx\cos ^{m+1}cx}{c(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}cx\cos ^{m}cx\;dx\qquad {\mbox{( }}m,n>0{\mbox{)}}\,\!}
∫
sin
n
c
x
cos
m
c
x
d
x
=
sin
n
+
1
c
x
cos
m
−
1
c
x
c
(
n
+
m
)
+
m
−
1
n
+
m
∫
sin
n
c
x
cos
m
−
2
c
x
d
x
(
m
,
n
>
0
)
{\displaystyle \int \sin ^{n}cx\cos ^{m}cx\;dx={\frac {\sin ^{n+1}cx\cos ^{m-1}cx}{c(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}cx\cos ^{m-2}cx\;dx\qquad {\mbox{( }}m,n>0{\mbox{)}}\,\!}
∫
d
x
sin
c
x
cos
c
x
=
1
c
ln
|
tg
c
x
|
{\displaystyle \int {\frac {dx}{\sin cx\cos cx}}={\frac {1}{c}}\ln \left|\operatorname {tg} cx\right|}
∫
d
x
sin
c
x
cos
n
c
x
=
1
c
(
n
−
1
)
cos
n
−
1
c
x
+
∫
d
x
sin
c
x
cos
n
−
2
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {dx}{\sin cx\cos ^{n}cx}}={\frac {1}{c(n-1)\cos ^{n-1}cx}}+\int {\frac {dx}{\sin cx\cos ^{n-2}cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
sin
n
c
x
cos
c
x
=
−
1
c
(
n
−
1
)
sin
n
−
1
c
x
+
∫
d
x
sin
n
−
2
c
x
cos
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {dx}{\sin ^{n}cx\cos cx}}=-{\frac {1}{c(n-1)\sin ^{n-1}cx}}+\int {\frac {dx}{\sin ^{n-2}cx\cos cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
sin
c
x
d
x
cos
n
c
x
=
1
c
(
n
−
1
)
cos
n
−
1
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {\sin cx\;dx}{\cos ^{n}cx}}={\frac {1}{c(n-1)\cos ^{n-1}cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
sin
2
c
x
d
x
cos
c
x
=
−
1
c
sin
c
x
+
1
c
ln
|
tg
(
π
4
+
c
x
2
)
|
{\displaystyle \int {\frac {\sin ^{2}cx\;dx}{\cos cx}}=-{\frac {1}{c}}\sin cx+{\frac {1}{c}}\ln \left|\operatorname {tg} \left({\frac {\pi }{4}}+{\frac {cx}{2}}\right)\right|}
∫
sin
2
c
x
d
x
cos
n
c
x
=
sin
c
x
c
(
n
−
1
)
cos
n
−
1
c
x
−
1
n
−
1
∫
d
x
cos
n
−
2
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{2}cx\;dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)\cos ^{n-1}cx}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
d
x
cos
c
x
=
−
sin
n
−
1
c
x
c
(
n
−
1
)
+
∫
sin
n
−
2
c
x
d
x
cos
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos cx}}=-{\frac {\sin ^{n-1}cx}{c(n-1)}}+\int {\frac {\sin ^{n-2}cx\;dx}{\cos cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
d
x
cos
m
c
x
=
sin
n
+
1
c
x
c
(
m
−
1
)
cos
m
−
1
c
x
−
n
−
m
+
2
m
−
1
∫
sin
n
c
x
d
x
cos
m
−
2
c
x
(
m
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}={\frac {\sin ^{n+1}cx}{c(m-1)\cos ^{m-1}cx}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}cx\;dx}{\cos ^{m-2}cx}}\qquad {\mbox{( }}m\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
d
x
cos
m
c
x
=
−
sin
n
−
1
c
x
c
(
n
−
m
)
cos
m
−
1
c
x
+
n
−
1
n
−
m
∫
sin
n
−
2
c
x
d
x
cos
m
c
x
(
m
≠
n
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}=-{\frac {\sin ^{n-1}cx}{c(n-m)\cos ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}cx\;dx}{\cos ^{m}cx}}\qquad {\mbox{( }}m\neq n{\mbox{)}}\,\!}
∫
sin
n
c
x
d
x
cos
m
c
x
=
sin
n
−
1
c
x
c
(
m
−
1
)
cos
m
−
1
c
x
−
n
−
1
m
−
1
∫
sin
n
−
1
c
x
d
x
cos
m
−
2
c
x
(
m
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}={\frac {\sin ^{n-1}cx}{c(m-1)\cos ^{m-1}cx}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-1}cx\;dx}{\cos ^{m-2}cx}}\qquad {\mbox{( }}m\neq 1{\mbox{)}}\,\!}
∫
cos
c
x
d
x
sin
n
c
x
=
−
1
c
(
n
−
1
)
sin
n
−
1
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {\cos cx\;dx}{\sin ^{n}cx}}=-{\frac {1}{c(n-1)\sin ^{n-1}cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
∫
cos
2
c
x
d
x
sin
c
x
=
1
c
(
cos
c
x
+
ln
|
tg
c
x
2
|
)
{\displaystyle \int {\frac {\cos ^{2}cx\;dx}{\sin cx}}={\frac {1}{c}}\left(\cos cx+\ln \left|\operatorname {tg} {\frac {cx}{2}}\right|\right)}
∫
cos
2
c
x
d
x
sin
n
c
x
=
−
1
n
−
1
(
cos
c
x
c
sin
n
−
1
c
x
)
+
∫
d
x
sin
n
−
2
c
x
)
(
n
≠
1
)
{\displaystyle \int {\frac {\cos ^{2}cx\;dx}{\sin ^{n}cx}}=-{\frac {1}{n-1}}\left({\frac {\cos cx}{c\sin ^{n-1}cx)}}+\int {\frac {dx}{\sin ^{n-2}cx}}\right)\qquad {\mbox{( }}n\neq 1{\mbox{)}}}
∫
cos
n
c
x
d
x
sin
m
c
x
=
−
cos
n
+
1
c
x
c
(
m
−
1
)
sin
m
−
1
c
x
−
n
−
m
−
2
m
−
1
∫
c
o
s
n
c
x
d
x
sin
m
−
2
c
x
(
m
≠
1
)
{\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}=-{\frac {\cos ^{n+1}cx}{c(m-1)\sin ^{m-1}cx}}-{\frac {n-m-2}{m-1}}\int {\frac {cos^{n}cx\;dx}{\sin ^{m-2}cx}}\qquad {\mbox{( }}m\neq 1{\mbox{)}}\,\!}
∫
cos
n
c
x
d
x
sin
m
c
x
=
cos
n
−
1
c
x
c
(
n
−
m
)
sin
m
−
1
c
x
+
n
−
1
n
−
m
∫
c
o
s
n
−
2
c
x
d
x
sin
m
c
x
(
m
≠
n
)
{\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}={\frac {\cos ^{n-1}cx}{c(n-m)\sin ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {cos^{n-2}cx\;dx}{\sin ^{m}cx}}\qquad {\mbox{( }}m\neq n{\mbox{)}}\,\!}
∫
cos
n
c
x
d
x
sin
m
c
x
=
−
cos
n
−
1
c
x
c
(
m
−
1
)
sin
m
−
1
c
x
−
n
−
1
m
−
1
∫
c
o
s
n
−
2
c
x
d
x
sin
m
−
2
c
x
(
m
≠
1
)
{\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}=-{\frac {\cos ^{n-1}cx}{c(m-1)\sin ^{m-1}cx}}-{\frac {n-1}{m-1}}\int {\frac {cos^{n-2}cx\;dx}{\sin ^{m-2}cx}}\qquad {\mbox{( }}m\neq 1{\mbox{)}}\,\!}
redaktə
∫
sin
c
x
tg
c
x
d
x
=
1
c
(
ln
|
sec
c
x
+
tg
c
x
|
−
sin
c
x
)
{\displaystyle \int \sin cx\operatorname {tg} cx\;dx={\frac {1}{c}}(\ln |\sec cx+\operatorname {tg} cx|-\sin cx)\,\!}
∫
tg
n
c
x
d
x
sin
2
c
x
=
1
c
(
n
−
1
)
tg
n
−
1
(
c
x
)
(
n
≠
1
)
{\displaystyle \int {\frac {\operatorname {tg} ^{n}cx\;dx}{\sin ^{2}cx}}={\frac {1}{c(n-1)}}\operatorname {tg} ^{n-1}(cx)\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
redaktə
∫
tg
n
c
x
d
x
cos
2
c
x
=
1
c
(
n
+
1
)
tg
n
+
1
c
x
(
n
≠
−
1
)
{\displaystyle \int {\frac {\operatorname {tg} ^{n}cx\;dx}{\cos ^{2}cx}}={\frac {1}{c(n+1)}}\operatorname {tg} ^{n+1}cx\qquad {\mbox{( }}n\neq -1{\mbox{)}}\,\!}
redaktə
∫
ctg
n
c
x
d
x
sin
2
c
x
=
1
c
(
n
+
1
)
ctg
n
+
1
c
x
(
n
≠
−
1
)
{\displaystyle \int {\frac {\operatorname {ctg} ^{n}cx\;dx}{\sin ^{2}cx}}={\frac {1}{c(n+1)}}\operatorname {ctg} ^{n+1}cx\qquad {\mbox{( }}n\neq -1{\mbox{)}}\,\!}
redaktə
∫
ctg
n
c
x
d
x
cos
2
c
x
=
1
c
(
1
−
n
)
tg
1
−
n
c
x
(
n
≠
1
)
{\displaystyle \int {\frac {\operatorname {ctg} ^{n}cx\;dx}{\cos ^{2}cx}}={\frac {1}{c(1-n)}}\operatorname {tg} ^{1-n}cx\qquad {\mbox{( }}n\neq 1{\mbox{)}}\,\!}
redaktə
∫
tg
m
(
c
x
)
ctg
n
(
c
x
)
d
x
=
1
c
(
m
+
n
−
1
)
tg
m
+
n
−
1
(
c
x
)
−
∫
tg
m
−
2
(
c
x
)
ctg
n
(
c
x
)
d
x
(
m
+
n
≠
1
)
{\displaystyle \int {\frac {\operatorname {tg} ^{m}(cx)}{\operatorname {ctg} ^{n}(cx)}}\;dx={\frac {1}{c(m+n-1)}}\operatorname {tg} ^{m+n-1}(cx)-\int {\frac {\operatorname {tg} ^{m-2}(cx)}{\operatorname {ctg} ^{n}(cx)}}\;dx\qquad {\mbox{( }}m+n\neq 1{\mbox{)}}\,\!}
redaktə
∫
−
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{)}}\,\!}
redaktə
∫
−
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
{\displaystyle \int _{-c}^{c}\cos {x}\;dx=2\int _{0}^{c}\cos {x}\;dx=2\int _{-c}^{0}\cos {x}\;dx\!}
∫
−
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
(voor
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{(voor }}n=1,3,5...{\mbox{)}}\,\!}
↑ Stewart, James. Calculus: Early Transcendentals, 6th Edition. Thomson: 2008
![{\displaystyle {\begin{aligned}\int x^{n}\sin cx\;dx&=n!\cdot \sin cx\left[{\frac {x^{n-1}}{c^{2}\cdot (n-1)!}}-{\frac {x^{n-3}}{c^{4}\cdot (n-3)!}}+{\frac {x^{n-5}}{c^{6}\cdot (n-5)!}}-...\right]-\\&-n!\cdot \cos cx\left[{\frac {x^{n}}{c\cdot n!}}-{\frac {x^{n-2}}{c^{3}\cdot (n-2)!}}+{\frac {x^{n-4}}{c^{5}\cdot (n-4)!}}-...\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7386e0080c6581e6c60f98d21aa1ea15e010715d)
