Question

If $f(x)={\sin }^{6}x+{\cos }^{6}x+2{\sin }^{3}x{\cos }^{3}x$, then $\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^{2}2x}{f(x)}dx=$

• A. $2$
• B. $\frac{2}{3}$
• C. $-\frac{2}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

Given, $f(x)={\sin }^{6}x+{\cos }^{6}x+2{\sin }^{3}x{\cos }^{3}x$
and $I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\sin }^{2}2x}{f(x)}dx$
$=\underset{0}{\overset{\pi 4}{\int }}\frac{4{\sin }^{2}x{\cos }^{2}x}{{\sin }^{6}x+{\cos }^{6}x+2{\sin }^{3}x{\cos }^{3}x}dx$
$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{4{\tan }^{2}x{\sec }^{2}x}{{\tan }^{6}x+2{\tan }^{3}x+1}dx$
Now, put ${\tan }^{3}x=t$, so at $x=0,t=0$ and at $x=\frac{\pi }{4}$
$t=1$ and $3{\tan }^{2}x{\sec }^{2}xdx=dt$
So, $I=\frac{4}{3}\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+2t+1}=\frac{4}{3}\underset{0}{\overset{1}{\int }}\frac{dt}{(t+1{)}^2}$
$={\frac{4}{3}\frac{-1}{(t+1)}|}_0^1=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{2}{3}$