How do you evaluate the integral `int sinx/(cosx + cos^2x) dx`?

Call the integral `I`. Note that the expression `cosx + cos^2x = cosx(1 + cosx)`. For partial fractions to work, expressions need to be factored the most possible.

`I = int sinx/(cosx(1 + cosx))`

To perform a partial fraction decomposition, we want to get rid of the trigonometric functions if possible. We can do this through a u-substitution. Let `u = cosx`. Then `du = -sinx dx` and `dx = (du)/(-sinx)`.

`I = int sinx/(u(1 + u)) * (du)/(-sinx)`

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

We're now going to use partial fraction decomposition to seperate integrals.

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

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

`Au + A + Bu = 1`

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

Now write a system of equations.

`{(A + B = 0), (A = 1):}`

This means that `A = 1` and `B = -1`.

The integral becomes.

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

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

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

`I=ln|cosx + 1| - ln|cosx| + C`

This can be simplified using `lna - lnb = ln(a/b)`.

`I = ln|(cosx + 1)/cosx|`

We can rewrite `1/cosx` as `secx`.

`I = ln|secx(cosx + 1)|`

`I = ln|1 + secx|`, since `secx` and `cosx` are reciprocals.

Hopefully this helps!

Let `I=intsinx/(cosx+cos^2x)dx.`

`"Subst. "u=cosx rArr du=-sinxdx," so, I=-int1/(u+u^2)du.`

We can integrate using Partial Factions, but, it is much simpler

without that.

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

`=-int{(u+1)/(u(u+1)}-u/(u(u+1))}du`

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

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

`=ln|(u+1)/u|=ln|u/u+1/u|`

Since, `u=cosx`, we have,

`I=ln|1+secx|+C.`

Enjoy Maths.!