Evaluate the following integral:`int_0^pixsinxcos^4x\ dx`

Explanation

Given:`int_0^pixsinxcos^4x\ dx`

`I=int_0^pixsinxcos^4x\ dx......(i)`

As we `int_b^a f(x)dx=int_b^a f(a-x)dx`

Then

`I=int_0^pi (pi-x)sinxcos^4xdx......(ii)`

On adding (i) and (ii) we get

`2I=int_0^pi (x+pi-x)sinxcos^4xdx`

`2I=pi int_0^pi sinxcos^4xdx`

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