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$$\int\frac1{(a^2\cos^2x+b^2\sin^2x)^2}\ dx$$

So this is the integral in question .

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

Another way of doing this is to substitute $b\tan x=a\tan t$.

So I was thinking if there any other ways to evaluate this integral as it seems to complicated as the given methods are very lengthy.

Any simpler/shorter method anyone could think of ?

Integreek
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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

2 Answers2

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Utilize \begin{align} I(a,b)=\int \frac{1}{ a^2\cos ^2 x + b^2 \sin ^2x}dx =\frac1{ab}\tan^{-1}\bigg({\frac ba}\tan x \bigg) \end{align} to compute \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)\\ &= \frac{1}{2a^2 b^2}\bigg[ (a^2+b^2)I(a,b) -\frac{a^2-b^2}{b^2\tan x+a^2\cot x}\bigg] \end{align}

Quanto
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0

Define $$P(x)=\frac{\sin x\cos x}{a^2\cos^2x+b^2\sin^2x}$$ $$\implies\int\frac{a^2\cos^2x-b^2\sin^2x}{\left(a^2\cos^2x+b^2\sin^2x\right)^2}\mathrm dx=P(x)+C\tag1$$

Rewriting the numerator of the integrand as $$a^2\cos^2x-b^2\sin^2x=\frac{a^2+b^2}{a^2-b^2}\left(a^2\cos^2x+b^2\sin^2x\right)+\frac{2a^2b^2}{b^2-a^2}$$

$(1)$ can be rewritten as:

$$\frac{a^2+b^2}{a^2-b^2}\int\frac{\mathrm dx}{a^2\cos^2x+b^2\sin^2x}+\frac{2a^2b^2}{b^2-a^2}\int\frac{\mathrm dx}{\left(a^2\cos^2x+b^2\sin^2x\right)^2}=P(x)+C$$

Using the result

$$\int\frac{\mathrm dx}{a^2\cos^2x+b^2\sin^2x}=\frac1{ab}\arctan\left(\frac{b}a\tan x\right)+C$$

The desired integral is

$$\boxed{\int\frac{\mathrm dx}{\left(a^2\cos^2x+b^2\sin^2x\right)^2}=\frac{a^2+b^2}{2a^3b^3} \arctan\left(\frac{b}a\tan x\right)+\frac{\left(b^2-a^2\right)\sin x\cos x}{2a^2b^2\left(a^2\cos^2x+b^2\sin^2x\right)}+C}$$

Integreek
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