# Partial Fraction Integral - Trig with Substitution

• Sep 6th 2011, 10:18 PM
joatmon
Partial Fraction Integral - Trig with Substitution
$\displaystyle \int \frac{\cos x dx}{6 \sin^2 x + \sin x}$

$let u = \sin x; du = \cos x dx$

$\int \frac{du}{6u^2 + u}$

$\frac{Ax + B}{6u^2 + u}$

$Ax + B = 1$

This leaves me with two variables, but only one equation! What have I missed???

Thanks!
• Sep 6th 2011, 10:24 PM
pickslides
Re: Partial Fraction Integral - Trig with Substitution
Quote:

Originally Posted by joatmon
$\displaystyle \int \frac{\cos x dx}{6 \sin^2 x + \sin x}$

$let u = \sin x; du = \cos x dx$

$\int \frac{du}{6u^2 + u}$

From here $\int \frac{du}{6u^2 + u} = \int \frac{du}{u(6u + 1)}$ , now use some partial fractions?
• Sep 6th 2011, 10:27 PM
Prove It
Re: Partial Fraction Integral - Trig with Substitution
Quote:

Originally Posted by joatmon
$\displaystyle \int \frac{\cos x dx}{6 \sin^2 x + \sin x}$

$let u = \sin x; du = \cos x dx$

$\int \frac{du}{6u^2 + u}$

$\frac{Ax + B}{6u^2 + u}$

$Ax + B = 1$

This leaves me with two variables, but only one equation! What have I missed???

Thanks!

$\displaystyle \frac{1}{6u^2 + u} = \frac{1}{u(6u + 1)}$

Then

\displaystyle \begin{align*} \frac{A}{u} + \frac{B}{6u + 1} &= \frac{1}{u(6u + 1)} \\ \frac{A(6u + 1) + Bu}{u(6u + 1)} &= \frac{1}{u(6u + 1)} \\ A(6u + 1) + Bu &= 1 \\ 6Au + A + Bu &= 1 \\ (6A + B)u + A &= 0u + 1 \\ 6A + B &= 0 \textrm{ and }A = 1 \\ 6 + B &= 0 \\ B &= -6 \end{align*}

So $\displaystyle \frac{1}{u} - \frac{6}{6u + 1} = \frac{1}{u(6u + 1)}$.

Go from here.
• Sep 6th 2011, 11:12 PM
joatmon
Re: Partial Fraction Integral - Trig with Substitution
Thank you! It's been a long day...