# Thread: Partial Fraction Decomposition w/ Trig Functions and Substitution

1. ## Partial Fraction Decomposition w/ Trig Functions and Substitution

Hi,

Can someone help me with this problem?

integral (9sec(theta))/(1+sin(theta))

WolframAlpha told me to multiply the top and bottom by -cos^2(theta), then converted it into something else using u=sin(theta)... but I don't understand. (I've reached my daily limit of free step-by-step solutions, so it won't let me view it anymore.) Is there another way to do this problem?

2. ## Re: Partial Fraction Decomposition w/ Trig Functions and Substitution

I would multiply the integrand in fact by $\displaystyle 1=\frac{\cos^2(\theta)}{\cos^2(\theta)}$ to get:

$\displaystyle 9\int\frac{\cos( \theta)}{\cos^2( \theta)(1+\sin( \theta))}\,d \theta$

then use the subsiitution:

$\displaystyle u=\sin( \theta)\,\therefore\,du=\cos( \theta)\,d\theta$

to get:

$\displaystyle 9\int\frac{1}{(1-u)(1+u)^2}\,du$

and now use partial fractions to complete the job, back-substituting for $\displaystyle u$ at the end.

3. ## Re: Partial Fraction Decomposition w/ Trig Functions and Substitution

\displaystyle \displaystyle \begin{align*} \int{\frac{9\sec{(\theta)}}{1 + \sin{(\theta)}}\,d\theta} &= \int{\frac{9}{\cos{(\theta)}\left[ 1 + \sin{(\theta)} \right] } \,d\theta} \\ &= \int{ \frac{9\cos{(\theta)}}{\cos^2{(\theta)} \left[ 1 + \sin{(\theta)} \right] } \,d\theta } \\ &= \int{ \frac{9\cos{(\theta)}}{\left[ 1 - \sin^2{(\theta)} \right] \left[ 1 - \sin{(\theta)} \right] } \, d\theta } \end{align*}

Now make the substitution \displaystyle \displaystyle \begin{align*} u = \sin{(\theta)} \implies du = \cos{(\theta)}\,d\theta \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} \int{\frac{9\cos{(\theta)}}{\left[ 1 - \sin^2{(\theta)} \right] \left[ 1 - \sin{(\theta)} \right] } \,d\theta} &= \int{ \frac{9}{ \left( 1 - u^2 \right) \left( 1 - u \right) } \,du } \\ &= \int{ \frac{9}{ \left( 1 - u \right) \left( 1 + u \right) \left( 1 - u \right) } \,du } \\ &= \int{ \frac{9}{\left( 1 - u \right) ^2 \left( 1 + u \right)} \,du } \end{align*}

Now applying Partial Fractions:

\displaystyle \displaystyle \begin{align*} \frac{A}{1 - u} + \frac{B}{ \left( 1- u \right) ^2 } + \frac{C}{ 1 + u} &\equiv \frac{ 9 }{ \left( 1 - u \right) ^2 \left( 1 + u \right) } \\ \frac{A \left( 1 - u \right) \left( 1+ u \right) + B \left( 1 + u \right) + C \left( 1 - u \right) ^2}{ \left( 1 - u \right) ^2 \left( 1 + u \right) } &\equiv \frac{9}{ \left( 1 - u \right) ^2 \left( 1 + u \right) } \\ A \left( 1 - u \right) \left( 1 + u \right) + B \left( 1 + u \right) + C \left( 1 - u \right) ^2 &\equiv 9\end{align*}

Let \displaystyle \displaystyle \begin{align*} u = 1 \end{align*} and we find \displaystyle \displaystyle \begin{align*} 2B = 9 \implies B = \frac{9}{2} \end{align*}.

Let \displaystyle \displaystyle \begin{align*} u = -1 \end{align*} and we find \displaystyle \displaystyle \begin{align*} 4C = 9 \implies C = \frac{9}{4} \end{align*}

Let \displaystyle \displaystyle \begin{align*} B= \frac{9}{2}, C = \frac{9}{2}, u = 0 \end{align*} and we find \displaystyle \displaystyle \begin{align*} A + \frac{9}{2} + \frac{9}{4} = 9 \implies A = \frac{9}{4} \end{align*}. Therefore

\displaystyle \displaystyle \begin{align*} \int{\frac{9}{\left( 1- u \right) ^2 \left( 1 + u \right) } \,du} &= \int{ \frac{\frac{9}{4}}{1 - u} + \frac{\frac{9}{2}}{\left( 1- u\right) ^2 } + \frac{\frac{9}{4}}{1 + u} \, du } \\ &= -\frac{9}{4}\ln{ \left| 1 - u \right|} + \frac{9}{2 \left( 1 - u \right) } + \frac{9}{4} \ln{ \left| 1 + u \right| } + C \\ &= \frac{9}{4} \left( \ln{ \left| \frac{1 + u}{1 - u} \right| } + \frac{2}{1 - u} \right) + C \\ &= \frac{9}{2} \left[ \ln{\left| \frac{1 + \sin{(\theta)}}{1 - \sin{(\theta)}} \right| } + \frac{2}{1 - \sin{(\theta)}} \right] + C \end{align*}

4. ## Re: Partial Fraction Decomposition w/ Trig Functions and Substitution

Please note that in my hurry to post this I made a few sign errors, I'll fix it when I get the chance. Thanks Mark for pointing them out

5. ## Re: Partial Fraction Decomposition w/ Trig Functions and Substitution

Ah! I see now. Thanks!