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I'm trying to use this subtitution $t=\tan(x/2)$. But I don´t get anywhere. I've tried $t=\tan(x)$ too. Appreciate your help.

$$\int\dfrac{\sin^2x}{1+\sin^2x}\mathrm dx$$

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  • $\begingroup$ What do mean by "I don't get anywhere"? Have you got the equivalent form of the integral after the change of variables? $\endgroup$ – Mhenni Benghorbal May 3 '14 at 20:22
  • $\begingroup$ I think you mean the substitution $ x=\arctan(t/2) $. $\endgroup$ – Mhenni Benghorbal May 3 '14 at 20:46
  • $\begingroup$ God,I've forgotten all my basic inverse trigonometric substitution identities. I better dig out my old copy of Strang and review.That's what sucks about doing rigorous math. After awhile,without review, you forget all the plug and chug stuff............. $\endgroup$ – Mathemagician1234 May 4 '14 at 7:07
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You can use the substitution $x=\arctan(t/2)$ and you will need the identity

$$ \sin( \arctan(t/2) ) = \frac{t}{\sqrt{t^2+1}} $$

to reach the form

$$ I= \int \frac{t^2}{(t^2+2)(t^2+4)}dt. $$

I think you can finish it now!

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Note that $$\frac{\sin^2{x}}{1 + \sin^2{x}} = \frac{1 + \sin^2{x}}{1 + \sin^2{x}} - \frac{1}{1 + \sin^2{x}}$$

The first factor integrates to $x$. For the second integral, note that $$\frac{1}{1 + \sin^2{x}} = \frac{\sec^2{x}}{\tan^2{x} + \sec^2{x}} = \frac{\sec^2{x}}{2\tan^2{x} + 1}$$

Proceed using the substitution $u = \tan{x}$.

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  • $\begingroup$ Great answer-especially for a beginner! This is why it's critical to know your trigonometric identities when calculating integrals. $\endgroup$ – Mathemagician1234 May 4 '14 at 20:45
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Write the ratio as

$$1- \frac{\cos^2 x}{1+\sin^2 x},$$

and recall that the derivative of $\arctan x$ is $1/(1+x^2)$ to obtain

$$x- 2^{-1/2} \arctan\bigl( \sqrt{2} \tan x\bigr)$$

(plus a constant)

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  • $\begingroup$ i think there is a typo. $\endgroup$ – Airbag May 3 '14 at 20:44

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