Indefinite Integration
Trigonometric Substitution
Grade 12
Question:
<p>\(\int \frac{\sin 2x \cos 2x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx\) is equal to (JEE Main 2018)</p>
<p>(a) \(\frac{-1}{3(1 + \tan^3 x)} + C\)</p>
<p>(b) \(\frac{1}{3(1 + \tan^3 x)} + C\)</p>
<p>(c) \(\frac{-1}{3(1 + \cot^3 x)} + C\)</p>
<p>(d) \(\frac{1}{1 + \cot^3 x} + C\)</p>
Step-by-Step Solution
Key Concept: Factor the denominator to recognize a standard form, then use substitution with inverse trigonometric or tangent functions.
<p><strong>Solution:</strong> Factor the denominator: $\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x = (\sin^2 x + \cos^2 x)(\sin^3 x + \cos^3 x) = \sin^3 x + \cos^3 x$. Let $u = 1 + \tan^3 x$, then $du = 3\tan^2 x \sec^2 x dx$. The integral simplifies to $\frac{-1}{3(1 + \tan^3 x)} + C$.</p>
Correct Answer: A