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$$\int\frac{1}{a^2\cos^2(x) + b^2 \sin^2(x)} \mathrm{d}x$$ I’ve tried to express $\sin$ through $\cos$ in the denominator and vice versa but it didn’t simplify the task.

FD_bfa
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Sofia Ibragimova
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    It can be written as $\frac{1}{b^2} \int \frac{\sec^2(x)}{(a/b)^2 + \tan^2(x)} \mathrm{d}x$. Substitute $\tan x = t$ and you will obtain a standard integral. – sudeep5221 Feb 08 '23 at 18:51
  • Mathematica shows the result is $\frac{\tan ^{-1}\left(\frac{b \tan (x)}{a}\right)}{a b}$. You can verify the derivation. – 138 Aspen Apr 19 '23 at 08:38

1 Answers1

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

Dividing numerator and denominator by $\cos^2x$

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

$$ =\int\frac{\sec^2 x \,dx}{a^2 + b^2 \tan^2(x)}= $$

Let $\tan x=t \implies \sec^2 xdx=dt $. Then, $$\int\frac{dt}{a^2t^2 + b^2 }=\frac{1}{b^2}\int \frac{dt}{\left(\frac{a^2t^2}{b^2}\right)+1}=\frac{1}{b^2}\int \frac{dt}{\left(\frac{at}{b}\right)^2+1}$$

$$= \frac1{b^2}\,\frac ba\,\arctan\left(\frac{at}{b}\right)+c= \frac1{ab} \,\arctan\left(\frac{a\tan x}{b}\right)+c$$

Sebastiano
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