JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q75

← JEE Main 2014 · 9 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 75 of 90

$\displaystyle\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}\,dx$ is equal to :

\dfrac12\sin2x+c

-\dfrac12\sin2x+c

-\dfrac12\sin x+c

-\sin^2x+c

Answer and worked solution

Official answer: B

Factor the numerator:
$\sin^8x-\cos^8x=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)=(-\cos2x)(1-2\sin^2x\cos^2x).$
The integrand is $-\cos2x$ , so the integral is $-\frac12\sin2x+c$ .

Topics:Standard integrals Trigonometric ratios and identities

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 75