MHT CET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} \mathrm{~d} x=\)
- A \(\frac{1}{3}\)
- B \(\frac{-1}{3}\)
- C \(\frac{1}{6}\)
- D \(\frac{-1}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{6}\)
Step-by-step Solution
Detailed explanation
\(\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} \mathrm{~d} x = \int_0^{\frac{\pi}{4}} \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} \mathrm{~d} x\) Subst. \(u = 1 + \tan^3 x \Rightarrow \mathrm{d} u = 3 \tan^2 x \sec^2 x \mathrm{~d} x\). Limits \(1 \to 2\)
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