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sciencesolve eNotes educator| Certified Educator

Supposing that you want to solve the integral `int x sin x dx` , you may use parts such that:

`u = x =gt du = dx`

`dv = sin x dx =gt v = int sin x dx = -cos x`

`int x sin x dx = -x cos x - int -cosx dx`

`int x sin x dx = -x cos x + int cos x dx`

`int x sin x dx = -x cos x + sin x + C`

Hence, evaluating the given integral using parts yields `int x sin x dx = -x cos x + sin x + C.`

justaguide eNotes educator| Certified Educator

There seems to be a typo in the question and the integral actually required is `int sin^2 x dx`

`int sin^2 x dx`

Use `cos 2x = 1 - 2*sin^2x => sin^2x = (1 - cos 2x)/2`

=> `int (1 - cos 2x)/2 dx`

=> `(1/2)*int 1 dx - (1/2)*int cos 2x dx`

=> `x/2 - (sin 2x)/4 + C`

=> `(2x - sin 2x)/4 + C`

The integral `int sin^2 x dx = (2x - sin 2x)/4 + C`

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