How do you use partial fractions to find the integral int (sinx)/(cosx+cos^2x)dx?

Dec 21, 2016

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

Explanation:

We start with a substitution. Let $u = \cos x$. Then $\mathrm{du} = - \sin x \mathrm{dx} \to \mathrm{dx} = \frac{\mathrm{du}}{-} \sin x$.

$\implies \int \sin \frac{x}{u + {u}^{2}} \times \frac{\mathrm{du}}{-} \sin x$

$\implies \int - \frac{1}{u + {u}^{2}} \mathrm{du}$

We now factor the denominator to $u \left(1 + u\right)$. Let's now write $- \frac{1}{u + {u}^{2}}$ in partial fractions.

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

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

$A u + A + B u = - 1$

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

Now write a system of equations:

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

Solving, we get: $A = - 1$ and $B = 1$.

Thus, the partial fraction decomposition is $- \frac{1}{u} + \frac{1}{u + 1}$. The integral becomes $\int \left(\frac{1}{u + 1} - \frac{1}{u}\right) \mathrm{du}$.

We can now integrate using the rule $\int \left(\frac{1}{u}\right) \mathrm{du} = \ln | u | + C$.

$\implies \ln | u + 1 | - \ln | u | + C$

Finally, reinsert the value of $u$ to make the function defined by $x$ instead of $u$:

$\implies \ln | \cos x + 1 | - \ln | \cos x | + C$

Hopefully this helps!