Question:
The integral \(\int\frac{\sin^2x\cos^2x}{\left(\sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x\right)^2}dx\) is equal to
The integral \(\int\frac{\sin^2x\cos^2x}{\left(\sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x\right)^2}dx\) is equal to
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Factor the denominator as \(\sin^3x+\cos^3x\), divide numerator and denominator by \(\cos^6x\) and substitute \(t=1+\tan^3x\).
Updated On: Nov 11, 2026
- \(\frac{-1}{1+\cot^3x}+C\)
- \(\frac{1}{3\left(1+\tan^3x\right)}+C\)
- \(\frac{-1}{3\left(1+\tan^3x\right)}+C\)
- \(\frac{1}{1+\cot^3x}+C\)
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The Correct Option is C
Solution and Explanation
Denominator: \(\sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x=\sin^2x(\sin^3x+\cos^3x)+\cos^2x(\sin^3x+\cos^3x)=\sin^3x+\cos^3x\).
So the integrand is \(\frac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}\).
Divide numerator and denominator by \(\cos^6x\): \(\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}\).
Let \(t=1+\tan^3x\), \(dt=3\tan^2x\sec^2x\,dx\).
\(I=\frac13\int\frac{dt}{t^2}=-\frac{1}{3t}+C=\frac{-1}{3(1+\tan^3x)}+C\).
Answer: (3) \(\frac{-1}{3(1+\tan^3x)}+C\)
So the integrand is \(\frac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}\).
Divide numerator and denominator by \(\cos^6x\): \(\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}\).
Let \(t=1+\tan^3x\), \(dt=3\tan^2x\sec^2x\,dx\).
\(I=\frac13\int\frac{dt}{t^2}=-\frac{1}{3t}+C=\frac{-1}{3(1+\tan^3x)}+C\).
Answer: (3) \(\frac{-1}{3(1+\tan^3x)}+C\)
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