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Please Integrate $$ \int_0^\pi \frac 1{(a^2\sin^2x + b^2 \cos^2)^2}\, dx$$ between limits 0 and pi. I have tried it multiple times but i keep falling into the tan inverse trap.

martini
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Achintya
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1 Answers1

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$$\int\frac{dx}{a^2\sin^2x+b^2\cos^2x}=\frac1{ab}\int\frac{\frac a{b\cos^2x}}{1+\left(\frac ab\tan x)^2\right)}dx=\frac1{ab}\arctan\left(\frac ab\tan x\right)+C\ldots$$

The above's based on the basic

$$\int\frac{f'(x)}{1+f(x)^2}dx=\arctan f(x)+C$$

DonAntonio
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  • The denominator is squared as a whole. Hence the above substitution does not help. – Achintya Feb 12 '14 at 14:46
  • Yeah...too bad that when I addressed this question the question was unedited and pretty blurry. BTW, the answer above has no substitution at all, and the same algebraic trick shown there can probably help for the squared denominator... – DonAntonio Feb 12 '14 at 15:53
  • I tried that but it didn't help. – Achintya Feb 13 '14 at 04:12