`int sinx/(cosx+cos^2x) dx` Use substitution and partial fractions to find the indefinite integral

`intsin(x)/(cos(x)+cos^2(x))dx`

Apply integral substitution: `u=cos(x)`

`=>du=-sin(x)dx`

`=int1/(u+u^2)(-1)du`

Take the constant out,

`=-1int1/(u+u^2)du`

Now to compute the partial fraction expansion of a proper rational function, we have to factor out the denominator,

`=-1int1/(u(u+1))du`

Now let's create the partial fraction expansion,

`1/(u(u+1))=A/u+B/(u+1)`

Multiply the above equation by the denominator,

`=>1=A(u+1)+B(u)`

`1=Au+A+Bu`

`1=(A+B)u+A`

Equating the coefficients of the like terms,

`A+B=0`  ------------------(1)

`A=1`

Plug in the value of A in equation 1,

`1+B=0`

`=>B=-1`

Plug in the values of...

`intsin(x)/(cos(x)+cos^2(x))dx`

Apply integral substitution: `u=cos(x)`

`=>du=-sin(x)dx`

`=int1/(u+u^2)(-1)du`

Take the constant out,

`=-1int1/(u+u^2)du`

Now to compute the partial fraction expansion of a proper rational function, we have to factor out the denominator,

`=-1int1/(u(u+1))du`

Now let's create the partial fraction expansion,

`1/(u(u+1))=A/u+B/(u+1)`

Multiply the above equation by the denominator,

`=>1=A(u+1)+B(u)`

`1=Au+A+Bu`

`1=(A+B)u+A`

Equating the coefficients of the like terms,

`A+B=0`  ------------------(1)

`A=1`

Plug in the value of A in equation 1,

`1+B=0`

`=>B=-1`

Plug in the values of A and B in the partial fraction expansion,

`1/(u(u+1))=1/u+(-1)/(u+1)`

`=1/u-1/(u+1)`

`int1/(u(u+1))du=int(1/u-1/(u+1))du`

Apply the sum rule,

`=int1/udu-int1/(u+1)du` 

Now use the common integral:`int1/xdx=ln|x|`

`=ln|u|-ln|u+1|`

Substitute back `u=cos(x)`

`=ln|cos(x)|-ln|cos(x)+1|`

`intsin(x)/(cos(x)+cos^2(x))dx=-1{ln|cos|x|-ln|cos(x)+1|}`

Simplify and add a constant C to the solution,

`=ln|cos(x)+1|-ln|cos(x)|+C`