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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 3

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

    Q3.    \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

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We have                                         I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}                                             ..................................................................(i)

By using   :

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

We get,                    

                                                   I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}

 

or                                              I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}                                                   . ............................................................(ii) 

 Adding (i) and (ii), we get : 

                                               2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

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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