# integration by partial fractions

• May 23rd 2007, 08:03 PM
Befordxide
integration by partial fractions
How do I integrate this integral using partial fractions?

\int_ dx/( Asin[x] + Bcos[x])^2 where a =/ 0

Thanks
• May 24th 2007, 07:37 PM
ThePerfectHacker
Quote:

Originally Posted by Befordxide
How do I integrate this integral using partial fractions?

\int_ dx/( Asin[x] + Bcos[x])^2 where a =/ 0

Thanks

Say $A,B\not = 0$

$\int \frac{dx}{A\sin x+B\cos x}$

Write,

$\frac{1}{A\sin x + B\cos x} = \frac{1}{A\cdot \frac{2\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}} + B\cdot \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}} = \frac{\frac{1}{2}(1+\tan^2 \frac{x}{2})}{A\tan^2 \frac{x}{2} +\frac{1}{2}B(1-\tan^2 \frac{x}{2} ) }$

The substituion $t = \tan \frac{x}{2}$ with $t'= \frac{1}{2}(1+\tan^2 \frac{x}{2})$

Will convert the integral into,

$\int \frac{dt}{At^2+\frac{1}{2}(B-t^2)}$

Can you do it from there?
• May 25th 2007, 11:08 AM
Soroban
Hello, Befordxide!

I have a back-door approach . . .

Quote:

$\int \frac{dx}{(A\!\cdot\!\sin x + B\!\cdot\!\cos x )^2}$

In the denominator we have: . $A\!\cdot\!\sin x + B\!\cdot\!\cos x$

Multiply top and bottom by $\sqrt{A^2+B^2}$: . $\sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\sin x + \frac{B}{\sqrt{A^2+B^2}}\cos x\right)$

Let $\theta$ be an acute angle in a right triangle with: . $adj = A,\:opp = B$
. . Then: . $hyp = \sqrt{A^2+B^2}$. . . Note that: . $\tan\theta = \frac{B}{A}$
. . Hence: . $\cos\theta \,= \,\frac{A}{\sqrt{A^2+B^2}},\;\;\sin\theta \,= \,\frac{B}{\sqrt{A^2+B^2}}$

Then we have: . $\sqrt{A^2+B^2}\left(\cos\theta\!\cdot\!\sin x + \sin\theta\!\cdot\!\cos x\right) \;=\;\sqrt{A^2+B^2}\cdot\sin(x + \theta)$

The denominator becomes: . $(A^2+B^2)\sin^2(x + \theta)$

The integral becomes: . $\frac{1}{\sqrt{A^2+B^2}}\int\frac{dx}{\sin^2(x + \theta)} \;=\;\frac{1}{\sqrt{A^2+B^2}}\int\csc^2(x + \theta)\,dx$

Can you finish it now?