∫ 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{)}}\,\!}
∫ 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{)}}\,\!}
∫ 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{)}}\,\!}
∫ 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{)}}\,\!}
∫ 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{)}}\,\!}
∫ 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{)}}\,\!}
∫ 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{)}}\,\!}
∫ − 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{)}}\,\!}
∫ − 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)
