Question

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

• 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 $I=\int\limits_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} 2 x}{f(x)} d x$
$=\int\limits_{0}^{\pi 4} \frac{4 \sin ^{2} x \cos ^{2} x}{\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x} d x$
$=\int\limits_{0}^{\frac{\pi}{4}} \frac{4 \tan ^{2} x \sec ^{2} x}{\tan ^{6} x+2 \tan ^{3} x+1} d x$
Now, put $\tan ^{3} x=t$, so at $x=0, t=0$ and at $x=\frac{\pi}{4}$
$t=1$ and $3 \tan ^{2} x \sec ^{2} x d x=d t$
So, $I=\frac{4}{3} \int\limits_{0}^{1} \frac{d t}{t^{2}+2 t+1}=\frac{4}{3} \int\limits_{0}^{1} \frac{d t}{(t+1)^{2}}$
$=\left.\frac{4}{3} \frac{-1}{(t+1)}\right|_{0} ^{1}=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{2}{3}$