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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS3 Marks
Question
By using the properties of definite integral, evaluate the integral in Exercise:$ \int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\sin^{2}\text{x}\ \text{dx}$
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Answer

$ \text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\sin^{2}\text{x}\ \text{dx}=2\int\limits_{0}^{\frac{\pi}{2}}\sin^{2}\text{x}\ \text{dx}..\text{(i)}\Big[\because\int^{\text{a}}_{-\text{a}}\text{f}\text{(x)}\ \text{dx}=2\int^{\text{a}}_{0}\text{f}\text{(x)}\text{dx},$when$\ \text{f(x)}$ is even function$\Big]$ $\Rightarrow\ \ \text{I}=2\int\limits_{0}^{\frac{\pi}{2}}\sin^{2}\bigg(\frac{\pi}{2}-\text{x}\bigg)\text{dx}\ \ \ \Big[\because\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x)}\text{dx}=\Big]$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\cos^{2}\text{x}\ \text{dx}$
Adding eq. (i) and (ii),
$21=2\int^{\frac{\pi}{2}}\limits_{0}\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}1\text{dx}=2\text{(x)}^{\frac{\pi}{2}}_{0}=2.\frac{\pi}{2}=\pi$
$\Rightarrow\ \ \ \text{I}=\frac{\pi}{2}$

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