Matematické Fórum

stuart clark · 01. 11. 2014 07:58

Evaluation of Integral

$(a)\;\; \int \frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx\;\;\;\;\;\;\;\;\;\;(b)\;\; \int\frac{\cos^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx$

Bati · 01. 11. 2014 16:34

Use the change of variables ( $t=\cot^2{x}$ for the first or $t=\tan^2{x}$ for the second), then decompose the fractions to obtain simple integrals. Note that it suffices to compute only one of the integrals, since
$a^2I_{(a)}+b^2I_{(b)}=x$ .

stuart clark · 02. 11. 2014 14:09

Thanks ↑ Bati:, Got it

Given $\bf{\int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx = \frac{1}{a^2}\int \frac{a^2\sin^2 x}{a^2\sin^2x+b^2\cos^2 x}dx}$

Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{a^2\cos^2 x\;,}$ We Get

$\bf{=\frac{1}{a^2}\int\frac{\tan^2 x}{\tan^2 x+A}dx = B\int\frac{\left(\tan^2 x+A^2\right)-A^2}{\tan^2 x+A^2}dx}$

Where $\bf{A^2 = \frac{b^2}{a^2}\;,}$ and $\bf{B=\frac{1}{a^2}}$

So Integral is $\bf{=B.\int\frac{\tan^2 x+A^2}{\tan^2 x+A^2}dx-A^2.B\int\frac{1}{\tan^2 x+A^2}dx}$

Now Put $\bf{\tan x=t\;,}$ and $\bf{dx = \frac{1}{\sec^2 x}dt = \frac{1}{1+t^2}dt}$

So Integral Convert into $\bf{=B.x-B.A^2\int\frac{1}{(t^2+A^2)(1+t^2)}dt}$

So $\bf{=B.x-\frac{A^2\cdot B}{\left(A^2-1\right)}\{\frac{(t^2+A^2)-(t^2+1)}{(t^2+1)(t^2+A^2)}\}dt}$

So $\bf{=B\cdot x - \frac{A^2\cdot B}{\left(A^2-1\right)}\int \{\frac{1}{1+t^2}-\frac{1}{A^2+t^2}\}dt}$

So Integral is $\bf{=B\cdot x -\frac{A^2\cdot B}{A^2-1}\tan^{-1}(t)-\frac{A^2\cdot B}{A^2-1}\cdot \frac{1}{A}\tan^{-1}\left(\frac{t}{A}\right)}$

Now put $\bf{A=\frac{b^2}{a^2}}$ and $\bf{B=\frac{1}{a^2}}$ and $\bf{t=\tan x\;,}$ We get

$\bf{\int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx=\frac{x}{a^2}-\frac{b^4}{a^2\left(b^4-a^4\right)}\tan^{-1}\left(x\right)-\frac{b^4}{a^2\left(b^4-a^4\right)}\tan^{-1}\left(\frac{a^2\tan x}{b^2}\right)+\mathcal{C}}$

Bati · 02. 11. 2014 15:03

↑ stuart clark:
Seeing your rather lengthy solution, I realize I made a shameful mistake in my change of variables which made the integration too easy. I'm sorry for that and i'm glad you managed to come out with a correct solution anyway.

Matematické Fórum

#1 01. 11. 2014 07:58

Indefinite Integral

#2 01. 11. 2014 16:34

Re: Indefinite Integral

#3 02. 11. 2014 14:09

Re: Indefinite Integral

#4 02. 11. 2014 15:03

Re: Indefinite Integral

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