The value of \int^{\pi\/3}_{-\pi\/3} \frac{xsinx}{\cos^2x} dx is :

Correct Answer is : 4π3 - 2 log tan 5π12

{4\pi}/3 - 2 log tan \frac{5\pi}{12}

{4\pi}/3 + log tan \frac{5\pi}{12}

{4\pi}/3 - log tan \frac{5\pi}{12}

NIMCET 2015 Question Paper with solutions

Question : 11 of 120

Marks: +1, -0

The value of

∫↖{π\/3}↙{-π\/3}

{xsinx}/{cos^2x}

dx is :

Solution:

We have,

∫^{π/3}↙{-π/3} {x\sin x}/{\cos^2x}

dx
Let

l=∫^{π/3}↙{-π/3}{x\sin x}/{\cos^2x}dx

⇒l=2∫↙{0}↖{π/3}{x\sin x}/{\cos^{2}x}dx

[ ∵

{x\sin x }/{\cos^{2}x}

is an even function]

⇒l=2∫↙{0}↖{π/3}x\sec x\tan xdx

⇒l=2[x∫\sec x\tan x\,dx-∫{dx}/{dx}∫\sec x\tan x\,dx\,⋅dx]_{0}^{π/3}

l=2[[x\sec x]_{0}^{π/3}-[\log(\sec x+\tan x)]_{0}^{π/3}]

⇒l=2[{π}/{3}\sec{π}/{3}-\log(\sec{π}/{3}+\tan{π}/{3})]

⇒l={4π}/{3}-2\log[{1+\sin{π}/{3}}/{\cos{π}/{3}}]

⇒l={4π}/{3}-2\log[{(\cos{π}/{6}+\sin{π}/{6})^{2}}/{\cos^{2}{π}/{6}-\sin^{2}{π}/{6}}]

⇒l={4π}/{3}-2\log[{\cos{π}/{6}+\sin{π}/{6}}/{\cos{π}/{6}-\sin{π}/{6}}]

⇒l={4π}/{3}-2\log\tan({π}/{4}+\;\;{π}/{6})

⇒l={4π}/{3}-2\log\tan{5π}/{12}

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