# I stuck with int (sin^(2)x*cosx)/(sinx+cosx)dx

mlehuzzah | Certified Educator

`("sin"^2x "cos"x)/("sin" x + "cos" x)`

`=("sin"^2x "cos"x ("sin" x - "cos" x))/(("sin" x + "cos" x)("sin" x - "cos" x))`

`=("sin"^2x * "sin" x "cos"x - "sin"^2 x "cos"^2 x)/("sin"^2 x - "cos"^2 x)`

`=((1/2)(1-"cos" 2x)(1/2)"sin" 2x - (1/4)"sin"^2 2x)/(-"cos" 2x)`

`=(-1/4) ("sin" 2x - "sin" 2x "cos" 2x - "sin"^2 2x)/("cos" 2x)`

`=(-1/4) ("tan" 2x - "sin" 2x) + (1/4) ("sin"^2 2x)/("cos" 2x)`

`=(-1/4) ("tan" 2x - "sin" 2x) + (1/4) (1 - "cos"^2 2x)/("cos" 2x)`

`=(-1/4) ("tan" 2x - "sin" 2x) + (1/4) ("sec" 2x - "cos" 2x)`

And we have:

`int ("sin"^2x "cos"x)/("sin" x + "cos" x) dx`

`=(1/4) int (-"tan" 2x + "sec" 2x ) dx`

Now, let `u=2x` `du = 2 dx`, so `dx = (1/2) du`

`=(1/4) int (-"tan" u + "sec" u ) (1/2) du`

`=(1/8) int (-"tan" u + "sec" u ) du`

`=(1/8) ["ln" |"cos" u| + "ln"| "sec" u + "tan" u| ]+ C`

`=(1/8) "ln"| "cos" u ("sec" u + "tan" u)| + C`

`=(1/8) "ln"| 1 + "sin" u| + C`

`=(1/8) "ln"| 1 + "sin" 2x| + C`

superpozicija | Student

Final result should be 1/4 ln |sinx+cosx|-(1/4)cosx(sinx+cosx)+c

...but i don't know how to get that result.