$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$

Question

My Try:
$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$

$\int\frac{a^2\tan^2x+b^2}{a^4\tan^2x+b^4}dx$

$\int\frac{1}{b^2}\frac{\frac{a^2}{b^2}\tan^2x+1}{\frac{a^4}{b^4}\tan^2x+1}dx$

I do not know how to solve it further.

Write it as $nr=a(dr)+b\frac{d}{dx}(dr)+c$ where nr is numerator dr is denominator and a,b,c are constants whose values are found out by comparing the equation — Archis Welankar, Feb 16 '16 at 13:22

vnd · Accepted Answer · 2016-02-16T13:21:45.427

4

The integrand can be written as $\frac{1}{a^2+b^2}\left[1+\frac{2a^2b^2}{\left(a^4+b^4\right)+\left(b^4-a^4\right)\cos 2\theta}\right]$. It is easy from here.

edited Feb 16 '16 at 13:21

answered Feb 16 '16 at 13:15

vnd

1,507

how did you get this expression?@vnd – Brahmagupta Feb 16 '16 at 13:17
2

express the squared sines and cosines in terms of $\cos 2\theta$ and then simplify by dividing. – vnd Feb 16 '16 at 13:19

$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$

1 Answers1