JEE Main On 8 April 2017 Question 6
Question: The integral $ \int\limits _{\frac{\pi }{12}}^{\frac{\pi }{4}}{\frac{8\cos 2x}{{{(tanx+cosx)}^{3}}}}dx $ equals: [JEE Online 08-04-2017]
Options:
A) $ \frac{13}{256} $
B) $ \frac{15}{64} $
C) $ \frac{13}{32} $
D) $ \frac{15}{128} $
Show Answer
Answer:
Correct Answer: D
Solution:
= $\int\limits _{\frac{\pi }{12}}^{\frac{\pi }{4}} \frac{8\cos 2x}{{\tan x+\cot x}^3}dx$
$\int\limits _{\frac{\pi }{12}}^{\frac{\pi }{4}} \frac{8\cos 2x8{\sin}^3x{\cos}^3x}{({{\sin}^2 x+{\cos}^2 x})^3 8}dx$
$\int\limits _{\frac{\pi }{12}}^{\frac{\pi }{4}} \frac{8\cos 2x{\sin 2x}^3}{({{\sin}^2 x+{\cos}^2 x})^3 8}dx$
$\int\limits _{\frac{\pi }{12}}^{\frac{\pi }{4}} {\cos 2x{\sin 2x}^3}dx$
put $\sin 2x=t\to 2 \cos 2x.dx=dt$
when $x=\frac{\pi}{12}\to t=\frac{1}{2}$
and $x=\frac{\pi}{4}\to=1$
$\int\limits _{\frac{1}{2}}^{1} {\cos 2x{t}^3}\frac{dx}{2\cos 2x}$
$\int\limits _{\frac{1}{2}}^{1} {{t}^3}\frac{dx}{2}$
$\frac{1}{2} [\frac{{t}^4}{4}]_{\frac{1}{2}}^{1} $
$\frac{1}{8}[1-\frac{1}{16}]$
$\frac{15}{128}$
