Question

Evaluate: $\int \frac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx$

• A. $\ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+c$
• B. $\ln \left|\frac{x\sin x+\cos x}{x\cos x+\sin x}\right|+c$
• C. $\ln \left|\frac{x\sin x-\cos x}{x\cos x+\sin x}\right|+c$
• D. None of these

Answer 1

The correct option is A $\ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+c$

Let $I=\int \frac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}dx$

$=\int [\frac{x\cos x}{x\sin x+\cos x}+\frac{x\sin x}{x\cos x-\sin x}]dx$

$=\int \frac{x\cos x}{x\sin x+\cos x}dx+\int \frac{x\sin x}{x\cos x-\sin x}dx$

$=I_1+I_2$

Put $x\sin x+\cos x=t\Rightarrow \left(x\cos x+\sin x-\sin x\right)dx=dt\Rightarrow x\cos xdx=dt$

$\Rightarrow I_1=\int \frac{1}{t}dt=\ln t+c=\ln (x\sin x+\cos x)+c$

And put $x\cos x-\sin x=s\Rightarrow (-x\sin x+\cos x-\cos x)dx=ds\Rightarrow -x\sin xdx=ds$

$\Rightarrow I_2=\int \frac{1}{s}ds=\ln s+c=\ln (x\cos x-\sin x)+c$

$\therefore I=\ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+c$