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MHTCET 15 May 2024 Shift 2 Paper

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Question : 121 of 150

Marks: +1, -0

\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=

$\frac{1}{2}$
$\frac{-1}{2}$
$\frac{3}{2}$
$\frac{-3}{2}$

Solution:

Concept:

Simplify the integrand using the identity

\sqrt{1+\cos x}\sqrt{1-\cos x}=\sin x

, then substitute

t=1-\cos x

.

Explanation:

Multiply the numerator and denominator by

\sqrt{1-\cos x}

.

\frac{\sqrt{1+\cos x}}{(1-\cos x)^{5/2}} \times \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} = \frac{\sqrt{1-\cos^2 x}}{(1-\cos x)^3}

For

x \in [\pi/3, \pi/2]

, we have

\sin x \ge 0

, so

\sqrt{1-\cos^2 x} = \sin x

.

Hence,

I = \int_{\pi/3}^{\pi/2} \frac{\sin x}{(1-\cos x)^3} \, dx

Let

t = 1 - \cos x

. Then

dt = \sin x \, dx

.

New limits: when

x = \pi/3

,

t = 1/2

; when

x = \pi/2

,

t = 1

.

I = \int_{1/2}^{1} \frac{dt}{t^3} = \left[-\frac{1}{2t^2}\right]_{1/2}^{1}

I = -\frac{1}{2} + \frac{1}{2}(4) = \frac{3}{2}

Answer:

\frac{3}{2}

, which corresponds to option C.

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