# find the integral of sin(3/2)xcos(1/2)x dx the answer is -1/4(cos2x+ 2cosx) + c, but how do i get there?

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We have to find the integral of sin (3x/2) * cos (x/2) dx

Int [ sin (3x/2) * cos (x/2) dx]

use sin 3x = 3sinx - 4(sinx)^3

=> Int [(3sin(x/2) - 4(sin(x/2))^3)*cos(x/2) dx]

let u = sin x/2

du/dx = (1/2)*cos (x/2)

=> 2*du = cos(x/2) dx

=> Int [(3u - 4u^3)*2 du]

=> 2* Int [(3u du] - 2* Int[ (4u^3) du]

=> 2* 3u^2 / 2 - 2* 4*u^4 /4 + C

=> 3u^2 - 2*u^4 + C

substitute u = sin x/2

=> 3*(sin x/2)^2 - 2*(sin x/2)^4 + C

=> (sin x/2)^2[ 3 - 2(sin x/2)^2] + C

=> (1/2)(1 - cos x)[ 2 + cos x] + C

=> (1/2)(2 - cos x - (cos x)^2) + C

=> (1/2)(2 - cos x - (1/2 + (1/2)(cos 2x)) + C

=> 1 - (1/2)cos x - 1/4 - (1/4)(cos 2x) + C

=> 3/4 - (1/4)(2cos x + (cos 2x)) + C

The 3/4 is added with the constant C

=> (-1/4)(cos 2x + 2*cos x) + C

**The required integral is (-1/4)(cos 2x + 2*cos x) + C**