JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 15)
The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $$ is equal to
(where C is a constant of integration):
(where C is a constant of integration):
$$\sec x - {{x\tan x} \over {x\sin x + \cos x}} + C$$
$$\sec x + {{x\tan x} \over {x\sin x + \cos x}} + C$$
$$\tan x - {{x\sec x} \over {x\sin x + \cos x}} + C$$
$$\tan x + {{x\sec x} \over {x\sin x + \cos x}} + C$$
Explanation
$${\int {\left( {{x \over {x\sin x + \cos x}}} \right)} ^2}dx $$
$$= \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}} $$
= $${x \over {\cos x}}\left( { - {1 \over {x\sin x + \cos x}}} \right) + \int {\left( {{{\cos x + x\sin x} \over {{{\cos }^2}x}}} \right)} \left( {{1 \over {x\sin x + \cos x}}} \right)dx$$
$$ = - {{x\sec x} \over {x\sin x + \cos x}} + \int {{{\sec }^2}x\,dx} $$
$$ = - {{x\sec x} \over {x\sin x + \cos x}} + \tan x + C$$
$$= \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}} $$
= $${x \over {\cos x}}\left( { - {1 \over {x\sin x + \cos x}}} \right) + \int {\left( {{{\cos x + x\sin x} \over {{{\cos }^2}x}}} \right)} \left( {{1 \over {x\sin x + \cos x}}} \right)dx$$
$$ = - {{x\sec x} \over {x\sin x + \cos x}} + \int {{{\sec }^2}x\,dx} $$
$$ = - {{x\sec x} \over {x\sin x + \cos x}} + \tan x + C$$
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