The integral (x^2)/((x x+ x)^2),dx is equal to (where C is a constant of integration):

Question

The integral $\displaystyle\int\frac{x^2}{(x\sin x+\cos x)^2}\,dx$ is equal to (where C is a constant of integration):

Accepted Answer

Note $\dfrac{d}{dx}(x\sin x+\cos x)=x\cos x$. Write $x^2=x\cos x\cdot\dfrac{x}{\cos x}$ so
$$I=\int\frac{x\cos x}{(x\sin x+\cos x)^2}\cdot\frac{x}{\cos x}\,dx$$

Integrating by parts with $u=\dfrac{x}{\cos x}=x\sec x$ and $dv=\dfrac{x\cos x}{(x\sin x+\cos x)^2}dx$ (so $v=-\dfrac1{x\sin x+\cos x}$):
$$I=\frac{-x\sec x}{x\sin x+\cos x}-\int\frac{-1}{x\sin x+\cos x}\cdot\frac{d}{dx}(x\sec x)\,dx$$

$\dfrac{d}{dx}(x\sec x)=\sec x+x\sec x\tan x=\sec x\cdot\dfrac{\cos x+x\sin x}{\cos x}\cdot\dfrac1{\cos x}\cdots$, and simplifying gives
$$I=\frac{-x\sec x}{x\sin x+\cos x}+\int\sec^2x\,dx=\frac{-x\sec x}{x\sin x+\cos x}+\tan x+C$$

Solutions · 1