JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q74

← JEE Main 2016 · 10 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 74 of 90

The integral $\displaystyle\int\frac{dx}{(1+\sqrt x)\sqrt{x-x^2}}$ is equal to :

(where C is a constant of integration.)

-2\sqrt{\dfrac{1+\sqrt x}{1-\sqrt x}}+C

-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+C

-\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+C

2\sqrt{\dfrac{1+\sqrt x}{1-\sqrt x}}+C

Answer and worked solution

Official answer: B

Put $x=\cos^2\theta$ : $dx=-2\sin\theta\cos\theta\,d\theta$ and $\sqrt{x-x^2}=\sin\theta\cos\theta$ . The integral becomes $-\int\frac{2\,d\theta}{1+\cos\theta}=-2\tan\frac\theta2+C$ , and $\tan\frac\theta2=\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}$ .

Topics:Substitution

← Previous question Next question →

JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 74