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$$ \int \frac{1}{ (a^2 \cos ^2 x + b ^ 2 \sin ^2x) ^2} \ dx $$

So this is the question .

The solution given in book is to divide numerator and denominator by $\cos ^4x$ and then substitute $\tan x = t$ in the resulting integrand.

Other way of doing this was to substitute $b \tan x = a \tan t$.

So I was thinking is not there any other way to solve this as it seems to a complicated problem as the given methods are very lengthy while solving.

Any simpler\shorter method anyone could think of ?

Quanto
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    I think there is none. – EQJ Oct 05 '18 at 13:37
  • A more general case of your question has been answered here – logo Oct 06 '18 at 15:13
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    @logo That is very different . also even that is not answered well . –  Oct 06 '18 at 16:06
  • Would it help at all to re-express the denominator as $\left((a^2-b^2)\cos^2x+b^2\right)^2$ via the identity $\sin^2x=1-\cos^2x$? You could clean it up a little more by letting $a^2-b^2=\alpha$ and $b^2=\beta$, so you're just left with $$\int\frac{1}{\left(\alpha \cos^2x+\beta\right)^2},dx$$ You can also plug it into integral-calculator.com and see what happens. – Robert Howard Oct 26 '18 at 01:34

1 Answers1

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Note that \begin{align} I(a,b)&=\int \frac{1}{ a^2\cos ^2 x + b^2 \sin ^2x}dx\\ &=\int \frac{d(\tan x)}{ a^2+b^2\tan ^2 x } =\frac1{ab}\tan^{-1}\bigg({\frac ba}\tan x \bigg) \end{align} and \begin{align} &\int \frac{1}{ (a^2\cos ^2 x + b^2 \sin ^2x)^2}dx = -\frac12\left(\frac{I’_a}a+\frac{I’_b}b\right)\\ =&\ \frac1{2ab}\left(\frac1{a^2}+\frac1{b^2}\right)\tan^{-1}\left(\frac ba\tan x \right) + \frac12\left(\frac1{a^2}-\frac1{b^2}\right)\frac{\tan x}{b^2\tan^2x+a^2} \end{align}

Quanto
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