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Integrate $\dfrac{dx}{(a^2 \cos^2x+b^2 \sin^2x)^2}$.

I can go up to the residue formula like in this example here

but then I just can't end up with the result for when $n=2$. I keep messing up my math. Can someone please show me how to get to the answer when $n=2$? I think I've spent 3+ hours trying to get to it and no luck unfortunately. Thank you!

Community
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user229056
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  • Welcome to math stackexchange. It is helpful to give as much info as you can about your problem, and to be careful with the tags. For example, when you say "complex" in complex-analysis, do you mean the integration may happen in the complex plane (with imaginary numbers), or do you mean that you feel the integral is complicated? The tag is more for the first. Second, is this an indefinite integral? If not, what are the limits? Lastly, there is a nice typesetting guide: http://math.stackexchange.com/help/notation to help your "math text" look more beautiful (and readable). – TravisJ Apr 06 '15 at 02:25

2 Answers2

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By setting $x=\arctan t$ we have: $$I=\int\frac{dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}=\int\frac{1+t^2}{(a^2 t^2+b^2)^2}\,dt$$ and by setting $t=u\frac{b}{a}$ we get: $$ I = \frac{1}{ab^3}\int\frac{1+\frac{b}{a} u^2}{(1+u^2)^2}\,du $$ so it is enough to check that: $$\int \frac{du}{(1+u^2)^2} = \frac{1}{2}\left(\frac{u}{1+u^2}+\arctan u\right),\quad \int \frac{u^2\, du}{(1+u^2)^2} = \frac{1}{2}\left(\arctan u-\frac{u}{1+u^2}\right).$$

Jack D'Aurizio
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$$ \begin{aligned} & \quad \int\frac{d x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)^2} \\ & =\int \frac{\sec ^4 x}{\left(a^2 \tan ^2 x+b^2\right)^2} d x \\ & =\int \frac{1+t^2}{\left(a^2 t^2+b^2\right)^2} d t \\ & =\int \frac{1+t^2}{a^4 t^4+2 a^2 b^2t^2+b^4} d t \\ & =\int \frac{\frac{1}{t^2}+1}{a^4 t^2+\frac{b^4}{t^2}+2 a^2 b^2} d t \\ & =\frac{1}{2} \int \frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\left(a^2+\frac{b^2}{t^2}\right)+\left(\frac{1}{a^2} -\frac{1}{b}\right)\left(a^2-\frac{b^2}{t^2}\right)}{a^4 t^2+\frac{b^4}{t^2}+2 a^2 b^2} d t \\ & =\frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right) \int \frac{d\left(a^2 t-\frac{b^2}{t}\right)}{\left(a^2 t-\frac{b^2}{t}\right)^2+4 a^2 b^2}+\frac{1}{2}\left(\frac{1}{a^2}-\frac{1}{b^2}\right) \int \frac{d\left(a^2 t+\frac{b^2}{t}\right)}{\left(a^2 t+\frac{b^2}{t}\right)^2} \\ & = \frac{1}{4a b} \left(\frac{1}{a^2}+\frac{1}{b^2}\right) \tan ^{-1}\left(\frac{a^2 t-\frac{b^2}{t}}{2 a b}\right) -\frac{1}{2}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)\left(\frac{t}{a^2 t^3+b^2}\right)+C \\ & \end{aligned} $$

By the way, putting back the limits yields $$\int_0^{\frac{\pi}{2}} \frac{d x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)^2}= \frac{\pi}{4 a b}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=\frac{\pi}{4 a^3 b^3}\left(a^2+b^2\right) $$

Lai
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