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I need to evaluate $$ \int_0^1 \dfrac 1{(x^2+1)^2}. $$

I was absent for this lecture and I'm sort of lost on where to go.

I set $x = \tan (\theta)$ and $dx = \sec^2(\theta)\, d\theta$. And this implies that

$$x^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta).$$

Now I have the integral of $\dfrac {\sec^2(\theta)}{(\sec^2\theta)^2}\, d\theta.$

I don't think I've done this properly, but if you could explain the process of doing trig substitutions in general, I would be greatly appreciative.

Sartree
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    Honestly, if I were you, I would evaluate $I(a)=\displaystyle\int_0^1\dfrac{dx}{x^2+a}~,~$ which is trivial, and then I'd express the entire integral in terms of $I'(1)$. – Lucian Mar 06 '15 at 04:48
  • @Lucian: Normally in a first course on integration, differentiation under the integral sign is not taught. –  Mar 06 '15 at 05:13
  • Here's yet another way: http://math.stackexchange.com/questions/70974/lesser-known-integration-tricks/689932#689932 – Hans Lundmark Mar 06 '15 at 06:39
  • I don't think I fully understand these, but thank you for pointing these methods out. Differentiation under the integral sign looks pretty useful and might be something useful for me to teach myself. – Sartree Mar 06 '15 at 13:07

3 Answers3

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Your are on the right track, but be careful as you are doing definite integral, you need also to change the lower and upper limit.

Now $x = \tan \theta$. When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \frac{\pi}{4}$. So you have

$$\int_0^1 \frac{1}{(1+ x^2)^2} dx = \int_0^{\pi/4} \frac{\sec ^2 \theta}{(\sec ^2\theta)^2}d\theta = \int_0^{\pi/4} \frac{1}{\sec ^2\theta }d\theta = \int_0^{\pi/4} \cos^2\theta d\theta$$

Now use the half angle formula

$$\cos ^2\theta = \frac{1}{2} (1+ \cos(2\theta))$$

to get

$$\int_0^{\pi/4} \cos^2\theta d\theta = \frac{1}{2} \int_0^{\pi/4} \big(1 + \cos (2\theta)\big) d\theta.$$

Can you fill in the remaining steps?

  • 1/2 int (1 + Cos2theta) from 0 to pi/4 d theta, equals:

    1/2 times x + sin(2theta)/2, evaluate from 0 to pi/4

    1/2( pi/4 + 1/2) = (pi + 2)/8

     – Sartree Mar 06 '15 at 03:03
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You have almost done! What is $\dfrac{1}{\sec\theta}$? Is that equal to $\cos\theta$? So you need to take a definite integral of $\cos^2\theta$ on that interval. Moreover, I'd like to warn you for taking any substitution for definite integral so fast. Indeed, we should satisfy some important conditions to do any substitution easily. See Relation to the fundamental theorem of calculus again.

amWhy
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Mikasa
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make a change of variable $$x = \tan t, t = \tan^{-1} x \to dt = \frac{dx}{1+x^2} $$

$$\int_0^1 \frac{dx}{(1+x^2)^2} = \int_0^{\pi/4} \frac{dt}{1+ \tan^2 t} = \int_0^{\pi/4} \cos^2 t \,dt = \frac 12\int_0^{\pi/4} 1 + \cos 2t \, dt = \frac 12\left( \frac{\pi}{4} + \frac12\right)$$

abel
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  • Thank you very much, this was helpful in checking my answer after working it out using some of the hints. – Sartree Mar 06 '15 at 03:05