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

Feb 17, 2017

The integral equals $\ln | \sec x + 1 | + C$

#### Explanation:

Call the integral $I$. Note that the expression $\cos x + {\cos}^{2} x = \cos x \left(1 + \cos x\right)$. For partial fractions to work, expressions need to be factored the most possible.

$I = \int \sin \frac{x}{\cos x \left(1 + \cos x\right)}$

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 $u = \cos x$. Then $\mathrm{du} = - \sin x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{- \sin x}$.

$I = \int \sin \frac{x}{u \left(1 + u\right)} \cdot \frac{\mathrm{du}}{- \sin x}$

$I = - \int \frac{1}{u \left(1 + u\right)} \mathrm{du}$

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

$\frac{A}{u} + \frac{B}{u + 1} = \frac{1}{u \left(u + 1\right)}$

$A \left(u + 1\right) + B u = 1$

$A u + A + B u = 1$

$\left(A + B\right) u + A = 1$

Now write a system of equations.

$\left\{\begin{matrix}A + B = 0 \\ A = 1\end{matrix}\right.$

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

The integral becomes.

$I = - \int \frac{1}{u} - \frac{1}{u + 1} \mathrm{du}$

$I = - \int \frac{1}{u} + \int \frac{1}{u + 1} \mathrm{du}$

$I = \ln | u + 1 | - \ln | u | + C$

$I = \ln | \cos x + 1 | - \ln | \cos x | + C$

This can be simplified using $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$.

$I = \ln | \frac{\cos x + 1}{\cos} x |$

We can rewrite $\frac{1}{\cos} x$ as $\sec x$.

$I = \ln | \sec x \left(\cos x + 1\right) |$

$I = \ln | 1 + \sec x |$, since $\sec x$ and $\cos x$ are reciprocals.

Hopefully this helps!

Feb 17, 2017

$\ln | 1 + \sec x | + C .$

#### Explanation:

Let $I = \int \sin \frac{x}{\cos x + {\cos}^{2} x} \mathrm{dx} .$

$\text{Subst. "u=cosx rArr du=-sinxdx,} s o , I = - \int \frac{1}{u + {u}^{2}} \mathrm{du} .$

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

without that.

$I = - \int \frac{1}{u \left(1 + u\right)} \mathrm{du} = - \int \frac{\left(u + 1\right) - u}{u \left(u + 1\right)} \mathrm{du}$

$= - \int \left\{\frac{u + 1}{u \left(u + 1\right)} - \frac{u}{u \left(u + 1\right)}\right\} \mathrm{du}$

$= \int \frac{1}{u + 1} \mathrm{du} - \int \frac{1}{u} \mathrm{du}$

$= \ln | u + 1 | - \ln | u |$

$= \ln | \frac{u + 1}{u} | = \ln | \frac{u}{u} + \frac{1}{u} |$

Since, $u = \cos x$, we have,

$I = \ln | 1 + \sec x | + C .$

Enjoy Maths.!