Table of Integrals: Proofs and Examples
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Basic Integrals / Trigonometric Integrals / Exponential and Logarithmic Integrals / Hyperbolic Integrals / Integrals Involving \(a^2+ u^2\) / Integrals Involving \(u^2- a^2\) / Integrals Involving \(a^2- u^2\) / Integrals Involving \(2au – u^2\) / Integrals Involving \(a + bu\)
Basic Integrals
- \(\displaystyle\int u^n du=\frac{u^{n+1}}{n+1}+C\), where \(n≠-1\)
- \(\displaystyle\int \frac{du}{u}=\mathrm{ln}\left|u\right|+C\)
- \(\displaystyle\int e^u\,du=e^u+C\)
- \(\displaystyle\int a^u\,du=\frac{a^u}{\mathrm{ln}a}+C\)
- \(\displaystyle\int \mathrm{sin} (u)du=-\mathrm{cos} (u)+C\)
- \(\displaystyle\int \mathrm{cos} (u)du=\mathrm{sin} (u)+C\)
- \(\displaystyle\int \mathrm{sec} (2u)du=\mathrm{tan} (u)+C\)
- \(\displaystyle\int \mathrm{csc} (2u)du=-\mathrm{cot} (u)+C\)
- \(\displaystyle\int \mathrm{sec} (u)\mathrm{tan} (u)du=\mathrm{sec} (u)+C\)
- \(\displaystyle\int \mathrm{csc} (u)\mathrm{cot} (u)du=-\mathrm{csc} (u)+C\)
- \(\displaystyle\int \mathrm{tan} (u)du=\mathrm{ln}\left|\mathrm{sec} (u)\right|+C\)
- \(\displaystyle\int \mathrm{cot} (u)du=\mathrm{ln}\left|\mathrm{sin} (u)\right|+C\)
- \(\displaystyle\int \mathrm{sec} (u)du=\mathrm{ln}\left|\mathrm{sec} (u)+\mathrm{tan} (u)\right|+C\)
- \(\displaystyle\int \mathrm{csc} (u)du=\mathrm{ln}\left|\mathrm{csc} (u)-\mathrm{cot} (u)\right|+C\)
- \(\displaystyle\int \frac{du}{\sqrt{a^2-u^2}}=\mathrm{sin^{-1}} (u) a+C\)
- \(\displaystyle\int \frac{du}{a^2+u^2}=\frac{1}{a} \mathrm{tan^{-1}} (u) a+C\)
- \(\displaystyle\int \frac{du}{u\sqrt{\sqrt{u^2-a^2}}}=\frac{1}{a} \mathrm{sec^{-1}} (u) a+C\)
Trigonometric Integrals
- \(\displaystyle\int \mathrm{sin} (2u)du=\frac{1}{2}u-\frac{1}{4}\mathrm{sin} (2u)+C\)
- \(\displaystyle\int \mathrm{cos} (2u)du=\frac{1}{2}u+\frac{1}{4}\mathrm{sin} (2u)+C\)
- \(\displaystyle\int \mathrm{tan} (2u)du=\mathrm{tan} (u)-u+C\)
- \(\displaystyle\int \mathrm{cot} (2u)du=-\mathrm{cot} (u)-u+C\)
- \(\displaystyle\int \mathrm{sin} (3u)du=-\frac{1}{3}(2+\mathrm{sin} (2u))\mathrm{cos} (u)+C\)
- \(\displaystyle\int \mathrm{cos} (3u)du=\frac{1}{3}(2+\mathrm{cos} (2u))\mathrm{sin} (u)+C\)
- \(\displaystyle\int \mathrm{tan} (3u)du=\frac{1}{2}\mathrm{tan} (2u)+\mathrm{ln}\left|\mathrm{cos} (u)\right|+C\)
- \(\displaystyle\int \mathrm{cot} (3u)du=-\frac{1}{2}\mathrm{cot} (2u)-\mathrm{ln}\left|\mathrm{sin} (u)\right|+C\)
- \(\displaystyle\int \mathrm{sec} (3u)du=\frac{1}{2}\mathrm{sec} (u)\mathrm{tan} (u)+\frac{1}{2}\mathrm{ln}\left|\mathrm{sec} (u)+\mathrm{tan} (u)\right|+C\)
- \(\displaystyle\int \mathrm{csc} (3u)du=-\frac{1}{2}\mathrm{csc} (u)\mathrm{cot} (u)+\frac{1}{2}\mathrm{ln}\left|\mathrm{csc} (u)-\mathrm{cot} (u)\right|+C\)
- \(\displaystyle\int \mathrm{sin}^{n} (u)du=-\frac{1}{n} \mathrm{sin}^{n-1} (u)\mathrm{cos} (u)+ \frac{n-1}{n} \int \mathrm{sin}^{n-2} (u)du\)
- \(\displaystyle\int \mathrm{cos}^{n} (u)du=\frac{1}{n} \mathrm{cos}^{n-1} (u)\mathrm{sin} (u)+ \frac{n-1}{n} \int \mathrm{cos}^{n-2} (u)du\)
- \(\displaystyle\int \mathrm{tan}^n (u)du=\frac{1}{n-1}\mathrm{tan}^{n-1} (u)-\int \mathrm{tan}^{n-2} (u)du\)
- \(\displaystyle\int \mathrm{cot}^n (u)du=-\frac{1}{n-1}\mathrm{cot}^{n-1} (u)-\int \mathrm{cot}^{n-2} (u)du\)
- \(\displaystyle\int \mathrm{sec}^n (u)du=\frac{1}{n-1}\mathrm{tan} (u)\mathrm{sec}^{n-2} (u)+ \frac{n-2}{n-1} \int \mathrm{sec}^{n-2} (u)du\)
- \(\displaystyle\int \mathrm{csc}^n (u)du=-\frac{1}{n-1}\mathrm{cot} (u)\mathrm{csc}^{n-2} (u)+ \frac{n-2}{n-1} \int \mathrm{csc}^{n-2} (u)du\)
- \(\displaystyle\int \mathrm{sin} (au)\mathrm{sin} (bu)du=\frac{\mathrm{sin} ((a-b)u)}{2(a-b)} – \frac{\mathrm{sin} ((a+b)u)}{2(a+b)}+C\)
- \(\displaystyle\int \mathrm{cos} (au)\mathrm{cos} (bu)du=\frac{\mathrm{sin} ((a-b)u)}{2(a-b)} + \frac{\mathrm{sin} ((a+b)u)}{2(a+b)}+C\)
- \(\displaystyle\int \mathrm{sin} (au)\mathrm{cos} (bu)du=-\frac{\mathrm{cos} ((a-b)u)}{2(a-b)}-\frac{\mathrm{cos} ((a+b)u)}{2(a+b)}+C\)
- \(\displaystyle\int u\mathrm{sin} (u)du=\mathrm{sin} (u)-u\mathrm{cos} (u)+C\)
- \(\displaystyle\int u\mathrm{cos} (u)du=\mathrm{cos} (u)+u\mathrm{sin} (u)+C\)
- \(\displaystyle\int u^n\mathrm{sin} (u)du=-u^n\mathrm{cos} (u)+n\int u^{n-1} \mathrm{cos} (u)du\)
- \(\displaystyle\int u^n\mathrm{cos} (u)du=u^n\mathrm{sin} (u)-n\int u^{n-1} \mathrm{sin} (u)du\)
- \(\displaystyle\int \mathrm{sin}^{n} (u)\mathrm{cos}^{m} (u)du =-\frac{\mathrm{sin}^{n-1} (u)\mathrm{cos}^{m+1} (u)}{n+m} + \frac{n-1}{n+m} \int \mathrm{sin}^{n-2} (u)\mathrm{cos}^{m} (u)du\)
\(\displaystyle\int \mathrm{sin}^{n} (u)\mathrm{cos}^{m} (u)du = \frac{\mathrm{sin}^{n+1} (u)\mathrm{cos}^{m-1} (u)}{n+m} + \frac{m-1}{n+m} \int \mathrm{sin}^{n} (u)\mathrm{cos}^{m-2} (u)du\)
Exponential and Logarithmic Integrals
- \(\displaystyle\int ue^{au}du= \frac{1}{a^2}(au-1)e^{au}+C\)
- \(\displaystyle\int u^ne^{au}du=\frac{1}{a}u^ne^{au}-\frac{n}{a} \int u^{n-1}e^{au}du\)
- \(\displaystyle\int e^{au}\mathrm{sin} (bu)du=\frac{e^{au}}{a^2+b^2}(a\mathrm{sin} (bu)-b\mathrm{cos} (bu))+C\)
- \(\displaystyle\int e^{au}\mathrm{cos} (bu)du=\frac{e^{au}}{a^2+b^2}(a\mathrm{cos} (bu)+b\mathrm{sin} (bu))+C\)
- \(\displaystyle\int \mathrm{ln}(u)du=u\mathrm{ln}(u)-u+C\)
- \(\displaystyle\int u^n\mathrm{ln}(u)du=\frac{u^{n+1}}{(n+1)^2}[(n+1)\mathrm{ln}(u)-1]+C\)
- \(\displaystyle\int \frac{1}{u\mathrm{ln}(u)} du=\mathrm{ln}\left|\mathrm{ln}(u)\right|+C\)
Hyperbolic Integrals
- \(\displaystyle\int \mathrm{sinh} (u)du=\mathrm{cosh} (u)+C\)
- \(\displaystyle\int \mathrm{cosh} (u)du=\mathrm{sinh} (u)+C\)
- \(\displaystyle\int \mathrm{tanh} (u)du=\mathrm{ln}\mathrm{cosh} (u)+C\)
- \(\displaystyle\int \mathrm{coth} (u)du=\mathrm{ln}\left|\mathrm{sinh} (u)\right|+C\)
- \(\displaystyle\int \mathrm{sech} (u)du=\mathrm{tan}^{-1}\left|\mathrm{sinh} (u)\right|+C\)
- \(\displaystyle\int \mathrm{csch} (u)du=\mathrm{ln}\left|\mathrm{tanh} (\frac{1}{2}u)\right|+C\)
- \(\displaystyle\int \mathrm{sech} (2u)du=\mathrm{tanh} (u)+C\)
- \(\displaystyle\int \mathrm{csch} (2u)du=-\mathrm{coth} (u)+C\)
- \(\displaystyle\int \mathrm{sech} (u)\mathrm{tanh} (u)du=-\mathrm{sech} (u)+C\)
- \(\displaystyle\int \mathrm{csch} (u)\mathrm{coth} (u)du=-\mathrm{csch} (u)+C\)
Inverse Trigonometric Integrals
- \(\displaystyle\int \mathrm{sin^{-1}} (u)du=u\mathrm{sin^{-1}} (u)+\sqrt{1-u^2}+C\)
- \(\displaystyle\int \mathrm{cos^{-1}} (u)du=u\mathrm{cos^{-1}} (u)-\sqrt{1-u^2}+C\)
- \(\displaystyle\int \mathrm{tan^{-1}} (u)du=u\mathrm{tan^{-1}} (u)-\frac{1}{2}\mathrm{ln}(1+u^2)+C\)
- \(\displaystyle\int u\mathrm{sin^{-1}} (u)du=\frac{2u^2-1}{4}\mathrm{sin^{-1}} (u)+ \frac{u\sqrt{1-u^2}}{4}+C\)
- \(\displaystyle\int u\mathrm{cos^{-1}} (u)du=\frac{2u^2-1}{4}\mathrm{cos^{-1}} (u)- \frac{u\sqrt{1-u^2}}{4}+C\)
- \(\displaystyle\int u\mathrm{tan^{-1}} (u)du=\frac{u^2+1}{2}\mathrm{tan^{-1}} (u)-u 2+C\)
- \(\displaystyle\int u^n\mathrm{sin^{-1}} (u)du=\frac{1}{n+1}\left[u^{n+1}\mathrm{sin^{-1}} (u)-\int \frac{u^{n+1}du}{\sqrt{1-u^2}}\right]\), where \(n≠-1\)
- \(\displaystyle\int u^n\mathrm{cos^{-1}} (u)du=\frac{1}{n+1}\left[u^{n+1}\mathrm{cos^{-1}} (u)+ \int \frac{u^{n+1}du}{\sqrt{1-u^2}}\right]\), where \(n≠-1\)
- \(\displaystyle\int u^n\mathrm{tan^{-1}} (u)du=\frac{1}{n+1}\left[u^{n+1}\mathrm{tan^{-1}} (u)- \int \frac{u^{n+1}du}{1+u^2}\right]\), where \(n≠-1\)
Integrals Involving \(a^2+ u^2\), where \(a > 0\)
- \(\displaystyle\int \sqrt{a^2+u^2}du=\frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}ln(u+\sqrt{a^2+u^2})+C\)
- \(\displaystyle\int u^2\sqrt{a^2+u^2}du=\frac{u}{8}(a^2+2u^2)\sqrt{a^2+u^2}-\frac{a^4}{8}ln(u+\sqrt{a^2+u^2})+C\)
- \(\displaystyle\int \frac{\sqrt{a^2+u^2}}{u}du=\sqrt{a^2+u^2}-a\mathrm{ln}\left|\frac{a+\sqrt{a^2+u^2}}{u}\right|+C\)
- \(\displaystyle\int \frac{\sqrt{a^2+u^2}}{u^2}du=-\frac{\sqrt{a^2+u^2}}{u}+ln(u+\sqrt{a^2+u^2})+C\)
- \(\displaystyle\int \frac{du}{\sqrt{a^2+u^2}}=ln(u+\sqrt{a^2+u^2})+C\)
- \(\displaystyle\int \frac{u^2du}{\sqrt{a^2+u^2}}=\frac{u}{2}(\sqrt{a^2+u^2})-\frac{a^2}{2}ln(u+\sqrt{a^2+u^2})+C\)
- \(\displaystyle\int \frac{du}{u\sqrt{a^2+u^2}}=-\frac{1}{a}\mathrm{ln}\left|\frac{\sqrt{a^2+u^2}+a}{u}\right|+C\)
- \(\displaystyle\int \frac{du}{u^2\sqrt{a^2+u^2}}=-\frac{\sqrt{a^2+u^2}}{a^2u}+C\)
- \(\displaystyle\int \frac{du}{(a^2+u^2)3/2}=\frac{u}{a^2\sqrt{a^2+u^2}}+C\)
Integrals Involving \(u^2- a^2\), where \(a > 0\)
- \(\displaystyle\int \sqrt{u^2-a^2}du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\mathrm{ln}\left|u+\sqrt{u^2-a^2}\right|+C\)
- \(\displaystyle\int u^2\sqrt{u^2-a^2}du=\frac{u}{8}(2u^2-a^2)\sqrt{u^2-a^2}-\frac{a^4}{8}\mathrm{ln}\left|u+\sqrt{u^2-a^2}\right|+C\)
- \(\displaystyle\int \frac{\sqrt{u^2-a^2}}{u}du=\sqrt{u^2-a^2}-a\mathrm{cos}^{-1}\left(\frac{a}{|u|}\right)+C\)
- \(\displaystyle\int \frac{\sqrt{u^2-a^2}}{u^2}du=-\frac{\sqrt{u^2-a^2}}{u}+\mathrm{ln}\left|u+\sqrt{u^2-a^2}\right|+C\)
- \(\displaystyle\int \frac{du}{\sqrt{u^2-a^2}}=\mathrm{ln}\left|u+\sqrt{u^2-a^2}\right|+C\)
- \(\displaystyle\int \frac{u^2du}{\sqrt{u^2-a^2}} = \frac{u}{2} \sqrt{u^2-a^2} + \frac{a^2}{2} \mathrm{ln}\left|u+ \sqrt{u^2-a^2}\right|+C\)
- \(\displaystyle\int \frac{du}{{u^2}\sqrt{u^2-a^2}}=\frac{\sqrt{u^2-a^2}}{a^2u}+C\)
- \(\displaystyle\int \frac{du}{(u^2-a^2)3/2} =-\frac{u}{a^2\sqrt{u^2-a^2}}+C\)
Integrals Involving \(a^2- u^2\), where \(a > 0\)
- \(\displaystyle\int \sqrt{a^2-u^2}du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\mathrm{sin^{-1}} (\frac{u}{a})+C\)
- \(\displaystyle\int u^2\sqrt{a^2-u^2}du=\frac{u}{8}(2u^2-a^2)\sqrt{a^2-u^2}+\frac{a^4}{8}\mathrm{sin^{-1}} (\frac{u}{a})+C\)
- \(\displaystyle\int \frac{\sqrt{a^2-u^2}}{u}du=\sqrt{a^2-u^2}-a\mathrm{ln}\left|\frac{a+\sqrt{a^2-u^2}}{u}\right|+C\)
- \(\displaystyle\int \frac{\sqrt{a^2-u^2}}{u^2}du=-\frac{1}{u}\sqrt{a^2-u^2}-\mathrm{sin^{-1}} (\frac{u}{a})+C\)
- \(\displaystyle\int \frac{u^2du}{\sqrt{a^2-u^2}}=-\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\mathrm{sin^{-1}} (\frac{u}{a})+C\)
- \(\displaystyle\int \frac{du}{u\sqrt{a^2-u^2}}=-\frac{1}{a}\mathrm{ln}\left|\frac{a+\sqrt{a^2-u^2}}{u}\right|+C\)
- \(\displaystyle\int \frac{du}{u^2\sqrt{a^2-u^2}}=-\frac{1}{a^2u}\sqrt{a^2-u^2}+C\)
- \(\displaystyle\int (a^2-u^2)^{3/2}du=-\frac{u}{8}(2u^2-5a^2)\sqrt{a^2-u^2}+\frac{3a^4}{8}\mathrm{sin^{-1}} (\frac{u}{a})+C\)
- \(\displaystyle\int \frac{du}{(a^2-u^2)^{3/2}}=-\frac{u}{a^2\sqrt{a^2-u^2}}+C\)
Integrals Involving \(2au – u^2\), where \(a > 0\)
- \(\displaystyle\int \sqrt{2au-u^2}du=\frac{u-a}{2}\sqrt{2au-u^2}+\frac{a^2}{2}\mathrm{cos}^{-1}(\frac{a-u}{a})+C\)
- \(\displaystyle\int \frac{du}{\sqrt{2au-u^2}}=\mathrm{cos}^{-1}(\frac{a-u}{a})+C\)
- \(\displaystyle\int u\sqrt{2au-u^2}du=\frac{2u^2-au-3a^2}{6}\sqrt{2au-u^2}+\frac{a^3}{2}\mathrm{cos}^{-1}(\frac{a-u}{a})+C\)
- \(\displaystyle\int \frac{du}{u\sqrt{2au-u^2}}=-\frac{\sqrt{2au-u^2}}{au}+C\)
Integrals Involving \(a + bu\), where \(a ≠ 0\)
- \(\displaystyle\int \frac{u\,du}{a+bu}=\frac{1}{b^2}(a+bu-a\mathrm{ln}\left|a+bu\right|)+C\)
- \(\displaystyle\int \frac{u^2 du}{a+bu}=\frac{1}{2b^3} \left[(a+bu)^2-4a(a+bu)+2a^2\mathrm{ln}\left|a+bu\right|\right]+C\)
- \(\displaystyle\int \frac{du}{{u}(a+bu)}=\frac{1}{a}\mathrm{ln}\left|\frac{u}{a+bu}\right|+C\)
- \(\displaystyle\int \frac{du}{{u^2}(a+bu)}=-\frac{1}{au}+\frac{b}{a^2}\mathrm{ln}\left|\frac{a+bu}{u}\right|+C\)
- \(\displaystyle\int \frac{u\,du}{(a+bu)^2}=\frac{a}{b^2(a+bu)}+\frac{1}{b^2}\mathrm{ln}\left|a+bu\right|+C\)
- \(\displaystyle\int \frac{u\,du}{u(a+bu)^2}=\frac{1}{a(a+bu)}-\frac{1}{a^2}\mathrm{ln}\left|\frac{a+bu}{u}\right|+C\)
- \(\displaystyle\int \frac{u^2 du}{(a+bu)^2}=\frac{1}{b^3}\left(a+bu-\frac{a^2}{a+bu}-2a\mathrm{ln}\left|a+bu\right|\right)+C\)
- \(\displaystyle\int u\sqrt{a+bu}du=\frac{2}{15b^2}(3bu-2a)(a+bu)^{3/2}+C\)
- \(\displaystyle\int \frac{u\,du}{\sqrt{a+bu}}=\frac{2}{3b^2}(bu-2a)\sqrt{a+bu}+C\)
- \(\displaystyle\int \frac{u^2 du}{\sqrt{a+bu}}=\frac{2}{15b^2}(8a^2+3b2u^2-4abu)\sqrt{a+bu}+C\)
- \(\displaystyle\int \frac{du}{u\sqrt{a+bu}}= \frac{1}{\sqrt{a}}\mathrm{ln}\left|\frac{\sqrt{a+bu}-\sqrt{a}}{\sqrt{a+bu}+\sqrt{a}}\right|+C\), if \(a>0\)
\(\displaystyle\int \frac{du}{u\sqrt{a+bu}}= \frac{2}{\sqrt{-a}}\mathrm{tan}^{-1}\left(\sqrt{\frac{a+bu}{-a}}\right)+C\), if \(a<0\) - \(\displaystyle\int \frac{\sqrt{a+bu}}{u}du=2\sqrt{a+bu}+a\int \frac{du}{u\sqrt{a+bu}}\)
- \(\displaystyle\int \frac{\sqrt{a+bu}}{u^2}du=-\frac{\sqrt{a+bu}}{u}+\frac{b}{2}\int \frac{du}{u\sqrt{a+bu}}\)
- \(\displaystyle\int u^n\sqrt{a+bu}du=\frac{2}{b(2n+3)}\left[u^n(a+bu)^{3/2}-na\int u^{n-1}\sqrt{a+bu}du\right]\)
- \(\displaystyle\int \frac{u^n du}{\sqrt{a+bu}}=\frac{2u^n\sqrt{a+bu}}{b(2n+1)}-\frac{2na}{b(2n+1)}\int \frac{u^{n-1}du}{\sqrt{a+bu}}\)
- \(\displaystyle\int \frac{du}{u^n\sqrt{a+bu}}=-\frac{\sqrt{a+bu}}{a(n-1)u^{n-1}}-\frac{b(2n-3)}{2a(n-1)}\int \frac{du}{u^{n-1}\sqrt{a+bu}}\)