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$\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t=$
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1983 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2}$
We have,
$$
\begin{aligned}
I &=\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t \\
&=\int_{0}^{\pi / 2} \frac{2 \sin t \cos t}{\sin ^{4} t+\cos ^{4} t} d t \\
&=\int_{0}^{\pi / 2} \frac{2 \tan t \sec ^{2} t}{\left(\tan ^{2} t\right)^{2}+1} d t
\end{aligned}
$$
Put $\tan ^{2} t=x$
$$
\begin{aligned}
&\left(2 \tan t \sec ^{2} t\right) d t=d x \\
&\therefore \quad I=\int_{0}^{\infty} \frac{1}{1+x^{2}} d x=\left[\tan ^{-1} x\right]_{0}^{\infty}=\frac{\pi}{2}
\end{aligned}
$$
$$
\begin{aligned}
I &=\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t \\
&=\int_{0}^{\pi / 2} \frac{2 \sin t \cos t}{\sin ^{4} t+\cos ^{4} t} d t \\
&=\int_{0}^{\pi / 2} \frac{2 \tan t \sec ^{2} t}{\left(\tan ^{2} t\right)^{2}+1} d t
\end{aligned}
$$
Put $\tan ^{2} t=x$
$$
\begin{aligned}
&\left(2 \tan t \sec ^{2} t\right) d t=d x \\
&\therefore \quad I=\int_{0}^{\infty} \frac{1}{1+x^{2}} d x=\left[\tan ^{-1} x\right]_{0}^{\infty}=\frac{\pi}{2}
\end{aligned}
$$
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