Q112 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics
$\int \frac{\sin 2 x d x}{\sin ^4 x+\cos ^4 x}=\tan ^{-1}(f(x))+c$, then $f\left(\frac{\pi}{3}\right)=$
- A1
- B2
- Correct answerC3
- D$\frac{1}{3}$
Solution
Let
$$
\begin{aligned}
I & =\int \frac{\sin 2 x d x}{\sin ^4 x+\cos ^4 x} \\
& =\int \frac{2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
\end{aligned}
$$
Dividing by $\cos ^2 x$ in numerator and denominator, we get
$$
=\int \frac{2 \tan x \sec ^2 x}{1+\left(\tan ^2 x\right)^2} d x
$$
Again let $\tan ^2 x=t$
$$
\begin{aligned}
\therefore \quad I & =\int \frac{d t}{1+t^2} \\
& =\tan ^{-1} t+C \\
& =\tan ^{-1}\left(\tan ^2 x\right)+C
\end{aligned}
$$
By comparing with $\tan ^{-1} f(x)+C$, we get
$$
\begin{aligned}
f(x) & =\tan ^2 x \\
\therefore \quad f\left(\frac{\pi}{3}\right) & =\left\{\tan \left(\frac{\pi}{3}\right)\right\}^2=(\sqrt{3})^2=3 .
\end{aligned}
$$
Hence, the correct answer is option: 'C'.