MHT CET · Maths · Indefinite Integration
The integral \(\int \frac{\sin ^2 x \cos ^2 x d x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2}\) is equal to (where \(C\) is a constant of integration).
- A \(\frac{1}{3\left(1+\tan ^3 x\right)}+C\)
- B \(\frac{-1}{3\left(1+\tan ^3 x\right)}+C\)
- C \(\frac{-1}{1+\cot ^3 x}+C\)
- D \(\frac{1}{1+\cot ^3 x}+C\)
Answer & Solution
Correct Answer
(B) \(\frac{-1}{3\left(1+\tan ^3 x\right)}+C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\sin ^2 x \cdot \cos ^2 x \mathrm{~d} x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \)
\( \int \frac{\tan ^2 x \cdot \sec ^6 x \mathrm{~d} x}{\left(\tan ^5 x+\tan ^2 x+\tan ^3 x+1\right)^2} \quad \text { [dividing } N^r\) \(\text {and } D^r \text { by } \cos ^{10} x \text { ] } \)
\( \int \frac{\tan ^2 x\left(\tan ^2 x+1\right)^2 \sec ^2 x \mathrm{~d} x}{\left(\tan ^2 x+1\right)^2\left(\tan ^3 x+1\right)^2} \)
\( \int \frac{\tan ^2 x \sec ^2 x \mathrm{~d} x}{\left(\tan ^3 x+1\right)^2} \)
\( =\frac{-1}{3\left(\tan ^3 x+1\right)}+c\left[\text { Let } \tan ^3 x+1=t\right]\)
\( \int \frac{\tan ^2 x \cdot \sec ^6 x \mathrm{~d} x}{\left(\tan ^5 x+\tan ^2 x+\tan ^3 x+1\right)^2} \quad \text { [dividing } N^r\) \(\text {and } D^r \text { by } \cos ^{10} x \text { ] } \)
\( \int \frac{\tan ^2 x\left(\tan ^2 x+1\right)^2 \sec ^2 x \mathrm{~d} x}{\left(\tan ^2 x+1\right)^2\left(\tan ^3 x+1\right)^2} \)
\( \int \frac{\tan ^2 x \sec ^2 x \mathrm{~d} x}{\left(\tan ^3 x+1\right)^2} \)
\( =\frac{-1}{3\left(\tan ^3 x+1\right)}+c\left[\text { Let } \tan ^3 x+1=t\right]\)
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