JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q75

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MathematicsSingle correct+4 / -1Question 75 of 90

The integral $\displaystyle\int_{\pi/12}^{\pi/4}\frac{8\cos2x}{(\tan x+\cot x)^3}\,dx$ equals :

\dfrac{15}{128}

\dfrac{15}{64}

\dfrac{13}{32}

\dfrac{13}{256}

Answer and worked solution

Official answer: A

$\tan x+\cot x=\dfrac2{\sin2x}$ , so the integrand is $\cos2x\sin^32x$ and
$\left[\frac{\sin^42x}8\right]_{\pi/12}^{\pi/4}=\frac18\left(1-\frac1{16}\right)=\frac{15}{128}.$

Topics:Fundamental theorem of calculus Substitution

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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 75