Question

Evaluate $\int \frac{\cos x}{x}dx$

Answer 1

Given,

$\int \frac{\cos x}{x}$

let, $v^{\prime }=\cos x,u=\frac{1}{x}$

$v=\sin x,u^{\prime }=\ln x$

$I=uv-\int vdu$

$I=\left(\frac{1}{x}\right)\sin x-\int \sin x\ln x$

let $u=\ln x,v^{\prime }=\sin x$

$u^{\prime }=\frac{1}{x},v=-\cos x$

$I=\left(\frac{1}{x}\right)\sin x-\left[\ln x\right(-\cos x)-\int -\cos x\left(\frac{1}{x}\right)]$

$I=\left(\frac{1}{x}\right)\sin x+\left[\ln x\right(-\cos x)-I]$

$2I=\frac{1}{x}\sin x+\ln x\cos x$

$\therefore I=\frac{1}{2}[\frac{1}{x}\sin x+\ln x\cos x]$