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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$
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Answer

$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}\times\frac{\sqrt{1-\cos\text{x}}}{\sqrt{1-\cos\text{x}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1-\cos^2\text{x}}}{(1-\cos\text{x})^2}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sin\text{x}}{(1-\cos\text{x})^2}\text{ dx}$
Let $1-\cos\text{x}=\text{t},$ Then $\sin\text{x dx}=\text{dt}$
When $\text{x}=\frac{\pi}{3},\text{ t}=\frac{1}{2}$ and $\text{x}=\frac{\pi}{2},\text{ t}=1$
Therefore the integral becomes
$=\int_{1}^{\frac{1}{2}}\frac{\text{dt}}{\text{t}^2}$
$=\Big[-\frac{1}{\text{t}}\Big]^1_\frac{1}{2}$
$=-1+2$
$=1$

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