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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 10

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

    Q10.    \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

Answers (1)

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We have                     I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

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

or                                I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx                                         ..............................................................(i)

By using the identity  : 

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

We get   : 

                                   I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx

or                               I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx                                         ....................................................................(ii)

 

Adding (i) and (ii) we get :-  

                            

                                2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx

or                               I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]

or                               I\ =\ \frac{\Pi }{2}\log\frac{1}{2}

Posted by

Devendra Khairwa

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