# I stuck with int (sin^(2)x*cosx)/(sinx+cosx)dxFinal result should be 1/4 ln |sinx+cosx|-(1/4)cosx(sinx+cosx)+c ...but i don't know how to get that result.

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You should come up with the following substitution such that:

`tan(x/2) = t => 1/2 sec^2(x/2) dx = dt`

`sin x = (2t)/(1 + t^2)`

`cos x = (1 - t^2)/(1 + t^2)`

Changing the variable yields:

`int (((8t^2)(1 - t^2))/(1 + t^2)^3)/((2t + 1 - t^2)/(1 + t^2))((1/(1 + t^2))dt`

Reducing like terms yields:

`8 int t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3`

You should use partial fraction decomposition such that:

`t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3 = A/(sqrt2 +t - 1)+B/(sqrt2 -t + 1) + (Ct+D)/(1+t^2) + (Et+F)/(1+t^2)^2 + (Gt+H)/(1+t^2)^3`

`t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3 = (At+B)(1+t^2)^3+ (Ct+D)(2t+1-t^2)(1+t^2)^2 + (Et+F)(2t+1-t^2)(1+t^2) +(Gt+H)(2t+1-t^2)`

`t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3 = (At+B)(1 + 3t^2 + 3t^4 + t^6) + (Ct+D)(2t+1-t^2)(1+2t^2 + t^4) + (Et+F)(2t+1-t^2)` `(1+t^2) + (Gt+H)(2t+1-t^2) t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3 = At + B + 3At^3 + 3Bt^2 + 3At^5 + 3Bt^4 + At^7 + Bt^6 + ....`

``

Hence, using partial fraction decomposition yields:

`A=1/32 B=-1/32`

`C=D=-1/16`

`E=-1/4, F=2/4`

`G=1, H=-1`

`8 int t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3 = int dt/(32(sqrt2 +t - 1)) - int dt/(32(sqrt2 -t + 1)) - int (t + 1)/(16(t^2+1)) dt + int (2-t)/(4(t^2+1)^2) dt + int (t-1)/(2(t^2+1)^2) dt`

Using substitution to solve integrals yields:

`t^2 + 1 = u => 2tdt = du => tdt = (du)/2`

`8 int t^2(1-t^2)/(2t+1-t^2)(1+t^2)^3 = (1/4)ln|(sqrt2 +t - 1)/(sqrt2 -t +1)| - (1/2) arctan t + int (2-t)/(4(t^2+1)^2) dt + int (t-1)/(2(t^2+1)^2) dt`

`int (sin^2x*cosx)/(sinx+cosx)dx = (1/4)ln |sin x+cos x| - (sin 2x)/8 - (cos 2x)/8 + c`

**Hence, evaluating the given integral yields `int (sin^2x*cosx)/(sinx+cosx)dx = (1/4)ln |sin x+cos x| - (sin 2x)/8 - (cos 2x)/8 + c.` **