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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 2 root sinx /(rootsinx+root(cosx)

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

    Q2.    \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

Answers (1)

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We have                        I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx                                                                        ..........................................................................(i)

By using , 

                                    \ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,               

                                   I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin (\frac{\pi}{2}-x)}+ \sqrt{\cos (\frac{\pi}{2}-x)}}dx

 

or                                                                                           I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx                          .......................................................(ii)

Adding (i) and (ii), we get,   

       

                                       2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

or                                    2I\ =\ \int_0^\frac{\pi}{2}1.dx

 

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

Thus                                   I\ =\ \frac{\Pi }{4}

Posted by

Devendra Khairwa

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