\displaystyle \textbf{FUNDAMENTAL INTEGRATION FORMULAS}

\displaystyle \text{We know that}
\displaystyle \frac{d}{dx}\{\phi(x)\}=f(x)\ \Leftrightarrow\ \int f(x)\,dx=\phi(x)+C,

\displaystyle \text{Based upon this and various standard differentiation formulae, we obtain the} \\ \text{following integration formulae:}

\displaystyle \text{(i)}\ \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n,\ n\neq -1\ \Rightarrow\ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\ n\neq -1

\displaystyle \text{(ii)}\ \frac{d}{dx}(\log_e x)=\frac{1}{x}\ \Rightarrow\ \int \frac{1}{x}\,dx=\log_e|x|+C

\displaystyle \text{(iii)}\ \frac{d}{dx}(e^x)=e^x\ \Rightarrow\ \int e^x\,dx=e^x+C

\displaystyle \text{(iv)}\ \frac{d}{dx}\left(\frac{a^x}{\log_e a}\right)=a^x,\ a>0,\ a\neq 1\ \Rightarrow\ \int a^x\,dx=\frac{a^x}{\log_e a}+C

\displaystyle \text{(v)}\ \frac{d}{dx}(-\cos x)=\sin x\ \Rightarrow\ \int \sin x\,dx=-\cos x+C

\displaystyle \text{(vi)}\ \frac{d}{dx}(\sin x)=\cos x\ \Rightarrow\ \int \cos x\,dx=\sin x+C

\displaystyle \text{(vii)}\ \frac{d}{dx}(\tan x)=\sec^2 x\ \Rightarrow\ \int \sec^2 x\,dx=\tan x+C

\displaystyle \text{(viii)}\ \frac{d}{dx}(-\cot x)=\mathrm{cosec}^2 x\ \Rightarrow\ \int \mathrm{cosec}^2 x\,dx=-\cot x+C

\displaystyle \text{(ix)}\ \frac{d}{dx}(\sec x)=\sec x\tan x\ \Rightarrow\ \int \sec x\tan x\,dx=\sec x+C

\displaystyle \text{(x)}\ \frac{d}{dx}(-\mathrm{cosec} x)=\mathrm{cosec} x\cot x\ \Rightarrow\ \int \mathrm{cosec} x\cot x\,dx=-\mathrm{cosec} x+C

\displaystyle \text{(xi)}\ \frac{d}{dx}(\log \sin x)=\cot x\ \Rightarrow\ \int \cot x\,dx=\log|\sin x|+C

\displaystyle \text{(xii)}\ \frac{d}{dx}(-\log \cos x)=\tan x\ \Rightarrow\ \int \tan x\,dx=-\log|\cos x|+C

\displaystyle \text{(xiii)}\ \frac{d}{dx}(\log(\sec x+\tan x))=\sec x\ \Rightarrow\ \int \sec x\,dx=\log|\sec x+\tan x|+C

\displaystyle \text{(xiv)}\ \frac{d}{dx}(\log(\mathrm{cosec} x-\cot x))=\mathrm{cosec} x\ \Rightarrow\ \int \mathrm{cosec} x\,dx=\log|\mathrm{cosec} x-\cot x|+C

\displaystyle \text{(xv)}\ \frac{d}{dx}\left(\sin^{-1}\left(\frac{x}{a}\right)\right)=\frac{1}{\sqrt{a^2-x^2}}\ \Rightarrow\ \int \frac{1}{\sqrt{a^2-x^2}}\,dx=\sin^{-1}\left(\frac{x}{a}\right)+C

\displaystyle \text{(xvi)}\ \frac{d}{dx}\left(\cos^{-1}\left(\frac{x}{a}\right)\right)=-\frac{1}{\sqrt{a^2-x^2}}\ \Rightarrow\ \int -\frac{1}{\sqrt{a^2-x^2}}\,dx=\cos^{-1}\left(\frac{x}{a}\right)+C

\displaystyle \text{(xvii)}\ \frac{d}{dx}\left(\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\right)=\frac{1}{a^2+x^2}\ \Rightarrow\ \int \frac{1}{a^2+x^2}\,dx=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C

\displaystyle \text{(xviii)}\ \frac{d}{dx}\left(\frac{1}{a}\cot^{-1}\left(\frac{x}{a}\right)\right)=-\frac{1}{a^2+x^2}\ \Rightarrow\ \int -\frac{1}{a^2+x^2}\,dx=\frac{1}{a}\cot^{-1}\left(\frac{x}{a}\right)+C

\displaystyle \text{(xix)}\ \frac{d}{dx}\left(\frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right)\right)=\frac{1}{x\sqrt{x^2-a^2}}\ \Rightarrow\ \int \frac{1}{x\sqrt{x^2-a^2}}\,dx=\frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right)+C

\displaystyle \text{(xx)}\ \frac{d}{dx}\left(\frac{1}{a}\mathrm{cosec}^{-1}\left(\frac{x}{a}\right)\right)=-\frac{1}{x\sqrt{x^2-a^2}}\Rightarrow \int -\frac{1}{x\sqrt{x^2-a^2}}\,dx=\frac{1}{a}\mathrm{cosec}^{-1}\left(\frac{x}{a}\right)+C

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