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Choose the correct answer in Exercises 20 and 21.

    Q21.    The value of     \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx is

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best_answer

We have

                                             I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )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}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx

or                                        I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx                                         .............................................................................(ii)

Adding (i) and (ii), we get:

                         

                                      2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx

or                                    2I\ =\ \int_0^\frac{\pi}{2}\log1.dx

Thus                               I\ =\ 0

 

Posted by

Devendra Khairwa

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