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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

    Q11.    \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

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We have                    I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

We know that  sin2x  is an even function.   i.e.,   sin2(-x)    =   (-sinx)2   =  sin2x.

 Also,                         

                                      I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx

So,                                

                                     I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx

or                                        =\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0

or                                    I\ =\ \frac{\Pi }{2}

Posted by

Devendra Khairwa

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