JEE Main 2014 · 12 Apr · Shift 1 · Mathematics Q74

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MathematicsSingle correct+4 / -1Question 74 of 90

The integral $\displaystyle\int\frac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}dx$ is equal to :

\dfrac1{(1+\cot^3x)}+c

-\dfrac1{3(1+\tan^3x)}+c

\dfrac{\sin^3x}{(1+\cos^3x)}+c

-\dfrac{\cos^3x}{3(1+\sin^3x)}+c

Answer and worked solution

Official answer: B

Divide top and bottom by $\cos^6x$ :
$\int\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}dx.$
With $u=1+\tan^3x$ , $du=3\tan^2x\sec^2x\,dx$ :
$\frac13\int\frac{du}{u^2}=-\frac1{3(1+\tan^3x)}+c.$

Topics:Substitution

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JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 74