JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q74

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MathematicsSingle correct+4 / -1Question 74 of 90

The integral $\displaystyle\int\sqrt{1+2\cot x(\operatorname{cosec}x+\cot x)}\,dx$ $\left(0<x<\dfrac\pi2\right)$ is equal to :

(where C is a constant of integration)

4\log\left(\sin\dfrac x2\right)+C

2\log\left(\sin\dfrac x2\right)+C

2\log\left(\cos\dfrac x2\right)+C

4\log\left(\cos\dfrac x2\right)+C

Answer and worked solution

Official answer: B

$1+2\cot x\operatorname{cosec}x+2\cot^2x=(\operatorname{cosec}x+\cot x)^2$ , and $\operatorname{cosec}x+\cot x=\cot\frac x2$ (positive here):
$\int\cot\frac x2\,dx=2\log\left(\sin\frac x2\right)+C.$

Topics:Standard integrals Trigonometric ratios and identities

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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 74