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

Question

The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $$ is equal to

(where C is a constant of integration):

Accepted Answer

$${\int {\left( {{x \over {x\sin x + \cos x}}} \right)} ^2}dx $$

$$= \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}} $$

= $${x \over {\cos x}}\left( { - {1 \over {x\sin x + \cos x}}} \right) + \int {\left( {{{\cos x + x\sin x} \over {{{\cos }^2}x}}} \right)} \left( {{1 \over {x\sin x + \cos x}}} \right)dx$$

$$ =  - {{x\sec x} \over {x\sin x + \cos x}} + \int {{{\sec }^2}x\,dx} $$

$$ =  - {{x\sec x} \over {x\sin x + \cos x}} + \tan x + C$$

Solutions · 1