Question

How do you integrate $\int {\cos }^{2}xdx$ by integration by parts.

Answer 1

Step- 1: Assume the variables:

The given integration is,

$\int {\cos }^{2}xdx=\int \cos x.\cos xdx$

Let $u=\cos x$ and $v=\cos x$.

Step- 2: Calculate using integration by parts:

Integration by parts is given as,

$\int u.vdx=u\int vdx-\int \left(u\text{'}\int vdx\right)dx$

$\therefore I=\cos x\left(\sin x\right)-\int \left(-\sin x\right)\left(\sin x\right)dx$

$=\sin x\cos x+\int {\sin }^{2}xdx$ $\left[\because \int cosxdx=sinx;\frac{d}{dx}\left(cosx\right)=-sinx\right]$

$=\sin x\cos x+\int \left(1-{\cos }^{2}x\right)dx$ $\left[\because si{n}^{2}x=1-co{s}^{2}x\right]$

$=\sin x\cos x+\int dx-\int {\cos }^{2}xdx$

$\Rightarrow$ $I=\sin x\cos x+x-I$ $\left[\because I=\int co{s}^{2}xdx\right]$

$\Rightarrow$ $2I=\sin x\cos x+x$

$\Rightarrow$ $I=\frac{\sin x\cos x}{2}+\frac{x}{2}+c\mathbf{[}\mathbf{c}\mathbf{}\mathbf{\text{be the integration constant}}\mathbf{]}$

Hence, the required answer is $\frac{\mathbf{s}\mathbf{i}\mathbf{nx}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{sx}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{+}\mathit{c}$.