A) \[\pi /4\]
B) \[\pi /2\]
C) \[\pi \]
D) None of these
Correct Answer: C
Solution :
\[4\int_{0}^{\pi /4}{\frac{\sin 2\theta \,d\theta }{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }=4\int_{0}^{\pi /4}{\frac{2\sin \theta \cos \theta \,d\theta }{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }}}\] \[=4\int_{0}^{\pi /4}{\frac{2\tan \theta {{\sec }^{2}}\theta \,d\theta }{{{\tan }^{4}}\theta +1}}\] {Dividing numerator and denominator by \[{{\cos }^{4}}\theta \]} Now put \[{{\tan }^{2}}\theta =t\Rightarrow 2\tan \theta {{\sec }^{2}}\theta \,d\theta =dt\], then the reduced form is \[4\int_{0}^{1}{\frac{dt}{{{t}^{2}}+1}}=4[{{\tan }^{-1}}t]_{0}^{1}=4\left[ \frac{1}{4}\pi -0 \right]=\pi \].
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