Some integration examples

Useful identities

\begin{eqnarray*} \sin^2 x+\cos^2 x=1 && \tan^2 x+1=\sec^2 x \\ \sin 2x=2\sin x\cos x&& \cos 2x=2\cos^2 x-1=1-2\sin^2x\\ \cos(a+b)+\cos(a-b)=2\cos a\cos b && \sin(a+b)+\sin(a-b)=2\sin a\cos b\\ \int \cos x dx=\sin x+C&&\int \sin xdx=-\cos x+C\\ \int \sec^2 dx=\tan x + C&&\int \tan xdx=\ln|\sec x|+c\\ &&\\ \cosh^2 x - \sinh^2 x=1 && 1-\tanh^2 x=\mathrm{sech}^2 x \\ \sinh 2x=2\sinh x\cosh x&& \cosh 2x=2\cosh^2 x-1=2\sinh^2x+1\\ \int \cosh x dx=\sinh x+C&&\int \sinh xdx=\cosh x+C\\ \end{eqnarray*}

Examples

\begin{enumerate} \item Find $\displaystyle\int_0^\frac{\pi}{4} 2\sin^2 2x dx.$ We re-arrange the integrand using the formula $\cos 2x=1-2\sin^2x$. This is because we can integrate a double angle directly. \begin{eqnarray*} \int_0^\frac{\pi}{4} 2\sin^2 2x dx&=&\int_0^\frac{\pi}{4} (1 - \cos 4x ) dx\\ &=&\left[x-\frac{\sin 4x}{4}\right]_0^\frac{\pi}{4}\\ &=&\frac{\pi}{4}. \end{eqnarray*} \item Find $\displaystyle\int_0^\frac{\pi}{2} \sin^3 x dx.$ We re-arrange the integrand using the formula $\sin ^2x=1-\cos^2x$. \begin{eqnarray*} \int_0^\frac{\pi}{2} \sin^3 x dx&=&\int \sin x(1 - \cos^2x ) dx\\ &=&\int_0^\frac{\pi}{2} (\sin x - \sin x\cos^2x ) dx.\\ \end{eqnarray*} Now we observe that $-\sin x$ is the derivative of $\cos x$ so that the second term is a sight integral of the form $\int f'(x)f^n(x) dx$. \begin{eqnarray*} \mbox{Thus }\int_0^\frac{\pi}{2} \sin^3 x dx&=&\int_0^\frac{\pi}{2} (\sin x - \sin x\cos^2x ) dx\\ &=&\left[-\cos x + \frac{\cos^3 x}{3}\right]_0^\frac{\pi}{2}\\ &=&-\left(-1 +\frac{1}{3}\right)\\ &=&\frac{2}{3}. \end{eqnarray*} \item Find $\displaystyle\int_0^\frac{\pi}{3} 2\tan^3 x dx.$ We re-arrange the integrand using the formula $\tan^2 x=\sec^2 x-1$. \begin{eqnarray*} \int_0^\frac{\pi}{3} 2\tan^3 xdx&=& \int_0^\frac{\pi}{3} (2\tan x)(\sec^2x -1)dx\\ &=&\int_0^\frac{\pi}{3} (2\tan x\sec^2 x- 2\tan x )dx. \end{eqnarray*} Now we observe that $\sec^2 x$ is the derivative of $\tan x$ so that the first term is a sight integral of the form $\int f'(x)f^n(x) dx$. \begin{eqnarray*} \int_0^\frac{\pi}{3} 2\tan^3 xdx&=& \int_0^\frac{\pi}{3} (2\sec^2 x\tan x- 2\tan x) dx\\ &=& \left[\tan^2 x - 2\ln|\sec x|\right]_0^\frac{\pi}{3}\\ &=& 3-2\ln 2. \end{eqnarray*} \item Find $\displaystyle\int_0^\frac{\pi}{3} 2\sin 2x \cos 3x dx.$ We re-arrange the integrand using the formula $\sin(a+b)+\sin(a-b)=2\sin a\cos b$. \begin{eqnarray*} \int_0^\frac{\pi}{3} 2\sin 2x \cos 3x dx &=& \int_0^\frac{\pi}{3} (\sin(5x) + \sin(-x))dx\\ &=& \int_0^\frac{\pi}{3} (\sin(5x) - \sin(x))dx\\ &=& \left[\frac{-\cos(5x)}{5}+\cos x\right]_0^\frac{\pi}{3}\\ &=&-\frac{1}{10}+\frac{1}{2}+\frac{1}{5}-1=\frac{-2}{5}. \end{eqnarray*} \item Find $\displaystyle\int 2\sqrt{x^2-1}dx$, for $x\geq1$. It is often a good idea to try the substitutions $x=\cos u$ or $x=\cosh u$ for integrals containing the terms $\sqrt{1-x^2}$ or $\sqrt{x^2-1}$ respectively (where there is no factor $x$). Since $x\geq 1$ let $x=\cosh u$. Then $\sqrt{x^2-1}=\sqrt{\cosh^2 u-1}=\sqrt{\sinh^2u}=\sinh u$ and $\displaystyle\frac{dx}{du}=\sinh u$. \begin{eqnarray*} \int 2\sqrt{x^2-1}dx &=&\int 2\sinh^2 u du\\ &=&\int (\cosh 2u-1) du \mbox{ using the `double angle' formula}\\ &=&\frac{\sinh 2u}{2}-u+C\\ &=&\sinh u\cosh u -u +C\\ &=&x\sqrt{x^2-1} - \cosh^{-1}(x) +C. \end{eqnarray*} Note that the integral is only defined if $x\leq -1$ or $x\geq 1$. In the former case we would let $x=-\cosh u$ and then proceed in the same way. \end{enumerate}