使用三角函式的積分有幾種不同的型別。我將把它分成幾個不同的部分。那些涉及正弦、餘弦和正切的。然後我將涵蓋餘切、正割和餘割。然後我將涵蓋反函式、涉及e、ln的函式，最後是雙曲函式。
需要注意的是，這些函式中使用的變數用u表示。（參見替換部分）
∫ sin u d u = − cos u + C {\displaystyle \int \sin u\mathrm {d} u=-\cos u+C}
∫ cos u d u = sin u + C {\displaystyle \int \cos u\mathrm {d} u=\sin u+C}
∫ tan u d u = − ln | cos u | + C {\displaystyle \int \tan u\mathrm {d} u=-\ln \left\vert \cos u\right\vert +C}
∫ sin 2 u d u = 1 2 ( u − sin u cos u ) + C {\displaystyle \int \sin ^{2}u\mathrm {d} u={\frac {1}{2}}(u-\sin u\cos u)+C}
∫ cos 2 u d u = 1 2 ( u + sin u cos u ) + C {\displaystyle \int \cos ^{2}u\mathrm {d} u={\frac {1}{2}}(u+\sin u\cos u)+C}
∫ tan 2 u d u = − u + tan u + C {\displaystyle \int \tan ^{2}u\mathrm {d} u=-u+\tan u+C}
∫ sin k u d u = − sin k − 1 u cos u k + k − 1 k ∫ sin k − 2 u d u {\displaystyle \int \sin ^{k}u\mathrm {d} u=-{\frac {\sin ^{k-1}u\cos u}{k}}+{\frac {k-1}{k}}\int \sin ^{k-2}u\mathrm {d} u}
∫ cos k u d u = cos k − 1 u sin u k + k − 1 k ∫ cos k − 2 u d u {\displaystyle \int \cos ^{k}u\mathrm {d} u={\frac {\cos ^{k-1}u\sin u}{k}}+{\frac {k-1}{k}}\int \cos ^{k-2}u\mathrm {d} u}
∫ tan k u d u = cos k − 1 u sin u k + k − 1 k ∫ cos k − 2 u d u {\displaystyle \int \tan ^{k}u\mathrm {d} u={\frac {\cos ^{k-1}u\sin u}{k}}+{\frac {k-1}{k}}\int \cos ^{k-2}u\mathrm {d} u}
∫ u sin u d u = sin u − u cos u + C {\displaystyle \int u\sin u\mathrm {d} u=\sin u-u\cos u+C}
∫ u cos u d u = cos
∫ u k sin u d u = − u k cos u + k ∫ u k − 1 cos u d u {\displaystyle \int u^{k}\sin u\mathrm {d} u=-u^{k}\cos u+k\int u^{k-1}\cos u\mathrm {d} u}
∫ u k cos u d u = u k sin u − k ∫ u k − 1 sin u d u {\displaystyle \int u^{k}\cos u\mathrm {d} u=u^{k}\sin u-k\int u^{k-1}\sin u\mathrm {d} u}
∫ 1 1 ± sin u d u = tan u ± sec u + C {\displaystyle \int {\frac {1}{1\pm \sin u}}\mathrm {d} u=\tan u\pm \sec u+C}
∫ 1 1 ± cos u d u = − cot u ± csc u + C {\displaystyle \int {\frac {1}{1\pm \cos u}}\mathrm {d} u=-\cot u\pm \csc u+C}
∫ 1 1 ± tan u d u = 1 2 ( u ± ln | cos u ± sin u | ) + C {\displaystyle \int {\frac {1}{1\pm \tan u}}\mathrm {d} u={\frac {1}{2}}(u\pm \ln \left\vert \cos u\pm \sin u\right\vert )+C}
∫ 1 sin u cos u d u = ln | tan u | + C {\displaystyle \int {\frac {1}{\sin u\cos u}}\mathrm {d} u=\ln \left\vert \tan u\right\vert +C}
∫ cot u d u = ln | sin u | + C {\displaystyle \int \cot u\mathrm {d} u=\ln \left\vert \sin u\right\vert +C} ∫ sec u d u = ln | sec u + tan u | + C {\displaystyle \int \sec u\mathrm {d} u=\ln \left\vert \sec u+\tan u\right\vert +C}
∫ csc u d u = ln | csc u − cot u | + C {\displaystyle \int \csc u\mathrm {d} u=\ln \left\vert \csc u-\cot u\right\vert +C}
∫ cot 2 u d u = − u − cot u + C {\displaystyle \int \cot ^{2}u\mathrm {d} u=-u-\cot u+C}
∫ sec 2 u d u = tan u + C {\displaystyle \int \sec ^{2}u\mathrm {d} u=\tan u+C}
∫ csc 2 u d u = − cot u + C {\displaystyle \int \csc ^{2}u\mathrm {d} u=-\cot u+C}
∫ cot k u d u = − cot k − 1 u k − 1 ∫ cot k − 2 u d , k ≠ 1 {\displaystyle \int \cot ^{k}u\mathrm {d} u=-{\frac {\cot ^{k-1}u}{k-1}}\int \cot ^{k-2}u\mathrm {d} ,k\neq 1}
∫ sec k u d u = sec k − 2 u tan u k − 1 + k − 2 k − 1 ∫ sec k − 2 u d u , l ≠ 1 {\displaystyle \int \sec ^{k}u\mathrm {d} u={\frac {\sec ^{k-2}u\tan u}{k-1}}+{\frac {k-2}{k-1}}\int \sec ^{k-2}u\mathrm {d} u,l\neq 1}
∫ csc k u d u = − csc k − 2 u tan u k − 1 + k − 2 k − 1 ∫ csc k − 2 u d u , k / n e q 1 {\displaystyle \int \csc ^{k}u\mathrm {d} u=-{\frac {\csc ^{k-2}u\tan u}{k-1}}+{\frac {k-2}{k-1}}\int \csc ^{k-2}u\mathrm {d} u,k/neq1}
∫ 1 1 ± cot u d u = 1 2 ( u ∓ ln | sin u ± cos u | ) + C {\displaystyle \int {\frac {1}{1\pm \cot u}}\mathrm {d} u={\frac {1}{2}}(u\mp \ln \left\vert \sin u\pm \cos u\right\vert )+C}
∫ 1 1 ± sec u d u = u + cot u ∓ csc u + C {\displaystyle \int {\frac {1}{1\pm \sec u}}\mathrm {d} u=u+\cot u\mp \csc u+C}
∫ 1 1 ± csc d u = u − tan u ± sec u + C {\displaystyle \int {\frac {1}{1\pm \csc }}\mathrm {d} u=u-\tan u\pm \sec u+C}
∫ arcsin u d u = u arcsin u + 1 − u 2 + C {\displaystyle \int \arcsin u\mathrm {d} u=u\arcsin u+{\sqrt {1-u^{2}}}+C}
∫ arccos u d u = u arccos u − 1 − u 2 + C {\displaystyle \int \arccos u\mathrm {d} u=u\arccos u-{\sqrt {1-u^{2}}}+C}
∫ arctan u d u = u arctan u − ln 1 + u 2 + C {\displaystyle \int \arctan u\mathrm {d} u=u\arctan u-\ln {\sqrt {1+u^{2}}}+C}
∫ arccot u d u = u arccot u + ln 1 + u 2 + C {\displaystyle \int \operatorname {arccot} u\mathrm {d} u=u\operatorname {arccot} u+\ln {\sqrt {1+u^{2}}}+C}
∫ arcsec u d u = u arcsec u + ln | u + u 2 − 1 | + C {\displaystyle \int \operatorname {arcsec} u\mathrm {d} u=u\operatorname {arcsec} u+\ln \left\vert u+{\sqrt {u^{2}-1}}\right\vert +C}
∫ arccsc u d u = u arccsc u + ln | u + u 2 − 1 | + C {\displaystyle \int \operatorname {arccsc} u\mathrm {d} u=u\operatorname {arccsc} u+\ln \left\vert u+{\sqrt {u^{2}-1}}\right\vert +C}
∫ e u d u = e u + C {\displaystyle \int e^{u}\mathrm {d} u=e^{u}+C}
∫ u e u d u = ( u − 1 ) e u + C {\displaystyle \int ue^{u}\mathrm {d} u=(u-1)e^{u}+C}
∫ u k e u d u = k ∫ u k − 1 e u d u {\displaystyle \int u^{k}e^{u}\mathrm {d} u=k\int u^{k-1}e^{u}\mathrm {d} u}
∫ 1 1 + e u d u − u − ln ( 1 + e u ) + C {\displaystyle \int {\frac {1}{1+e^{u}}}\mathrm {d} u-u-\ln(1+e^{u})+C}
∫ e a u sin b u d u = e a u a 2 + b 2 ( a sin b u − b cos b u ) + C {\displaystyle \int e^{au}\sin bu\mathrm {d} u={\frac {e^{au}}{a^{2}+b^{2}}}(a\sin bu-b\cos bu)+C}
∫ e a u cos b u d u = e a u a 2 + b 2 ( a cos b u + b sin b u ) + C {\displaystyle \int e^{au}\cos bu\mathrm {d} u={\frac {e^{au}}{a^{2}+b^{2}}}(a\cos bu+b\sin bu)+C}
∫ ] l n u d u = u ( − 1 + ln u ) + C {\displaystyle \int ]lnu\mathrm {d} u=u(-1+\ln u)+C}
∫ u ln u d u = u 2 4 ( − 1 + 2 ln u ) + C {\displaystyle \int u\ln u\mathrm {d} u={\frac {u^{2}}{4}}(-1+2\ln u)+C}
∫ u k ln u d u = e a u a 2 + b 2 ( a cos b u + b sin b u ) + C {\displaystyle \int u^{k}\ln u\mathrm {d} u={\frac {e^{au}}{a^{2}+b^{2}}}(a\cos bu+b\sin bu)+C}
∫ ( ln u ) 2 d u = u [ 2 − 2 ln u + ( ln u ) 2 ] + C {\displaystyle \int (\ln u)^{2}\mathrm {d} u=u[2-2\ln u+(\ln u)^{2}]+C}
∫ ( ln u ) k d u = u ( ln u ) k − k ∫ ( ln u ) k − 1 d u {\displaystyle \int (\ln u)^{k}\mathrm {d} u=u(\ln u)^{k}-k\int (\ln u)^{k-1}\mathrm {d} u}
∫ cosh u d u = sinh u + C {\displaystyle \int \cosh u\mathrm {d} u=\sinh u+C}
∫ sinh u d u = cosh u + C {\displaystyle \int \sinh u\mathrm {d} u=\cosh u+C}
∫ sech 2 u d u = tanh u + C {\displaystyle \int \operatorname {sech} ^{2}u\mathrm {d} u=\tanh u+C}
∫ csch 2 u d u = − coth u + C {\displaystyle \int \operatorname {csch} ^{2}u\mathrm {d} u=-\coth u+C}
∫ sech u tan u d u = − sech u + C {\displaystyle \int \operatorname {sech} u\tan u\mathrm {d} u=-\operatorname {sech} u+C}
∫ csch u coth u d u = − csch u + C {\displaystyle \int \operatorname {csch} u\coth u\mathrm {d} u=-\operatorname {csch} u+C}
![{\displaystyle \int ]lnu\mathrm {d} u=u(-1+\ln u)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a83b3d12d35a5b868c565cbb81f21441cfe9f2e)
![{\displaystyle \int (\ln u)^{2}\mathrm {d} u=u[2-2\ln u+(\ln u)^{2}]+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a06ecae7479c489698af96690a749d9c6c5dcb03)
