MHT CET · Maths · Definite Integration
\(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=\)
- A \(\frac{1}{2}\)
- B \(\frac{-1}{2}\)
- C \(\frac{3}{2}\)
- D \(\frac{-3}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
Let \(\begin{aligned} I & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x \\ & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} \times \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} d x \\ & =\int_{\pi / 3}^{\pi / 2} \frac{\sin x}{(1-\cos x)^3} d x\end{aligned}\)
\(\begin{array}{ll} & \text { Put } 1-\cos x=\mathrm{t} \\ & \Rightarrow \sin x \mathrm{~d} x=\mathrm{dt} \\ \therefore \quad & \mathrm{I}=\int_{1 / 2}^1 \frac{\mathrm{dt}}{\mathrm{t}^3}=\left[\frac{\mathrm{t}^{-2}}{-2}\right]_{1 / 2}^1=\frac{3}{2}\end{array}\)
\(\begin{array}{ll} & \text { Put } 1-\cos x=\mathrm{t} \\ & \Rightarrow \sin x \mathrm{~d} x=\mathrm{dt} \\ \therefore \quad & \mathrm{I}=\int_{1 / 2}^1 \frac{\mathrm{dt}}{\mathrm{t}^3}=\left[\frac{\mathrm{t}^{-2}}{-2}\right]_{1 / 2}^1=\frac{3}{2}\end{array}\)
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