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Question

By using the properties of definite integrals, evaluate the integral   $$\displaystyle \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\sin^2 x dx$$


Solution

Let $$\displaystyle I=\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\sin^2 x dx$$
As $$\sin^2(-x)=(\sin (-x))^2=(- \sin x)^2=\sin^2x$$,
Therefore, $$\sin^2x$$ is an even functions.
It is known that if $$f(x)$$ is an even function, then $$\displaystyle\int_{-a}^af(x)dx=2\int_0^af(x)dx$$
$$\displaystyle\therefore I=2\int_0^{\frac {\pi}{2}}\sin^2 x dx$$
$$\displaystyle=2\int_0^{\frac {\pi}{2}}\frac {1-\cos 2x}{2}dx$$
$$\displaystyle=\int_0^{\frac {\pi}{2}}(1-\cos 2x)dx$$
$$\displaystyle=\left [x-\frac {\sin 2x}{2}\right ]_0^{\frac {\pi}{2}}=\frac {\pi}{2}$$

Mathematics
RS Agarwal
Standard XII

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