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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 15

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

    Q15.    \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx

Answers (1)

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We have                    I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx                                  ................................................................(i)

 

By using the property :-  

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

We get ,                

                                          I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx

or                                      I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx                                 ......................................................................(ii)

 

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

 

                                        2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx

Thus                                 I   =   0

Posted by

Devendra Khairwa

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