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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 4

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

    Q4.    \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}

Answers (1)

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We have                                         I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}                                             ..................................................................(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{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}

    

or                                                I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x}                                                     . ............................................................(ii) 

 Adding (i) and (ii), we get : 

                                               2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}

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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