# Thread: trig substitution

1. ## trig substitution

Good afternoon all,

I am looking at what I hope is my last trig substitution question:

$\displaystyle \int_{0}^{1}{\frac{1}{x^2+2x+2}dx$

Through completing the square I have:

$\displaystyle \int_{0}^{1}{\frac{1}{(x+1)^2+1}dx$

My chosen substitution is:

$\displaystyle x = \sqrt{1}tan\theta = tan\theta$
and
$\displaystyle x = \sqrt{1}sec^2\theta d\theta = sec^2\theta d\theta$

Using this substitution I get:

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

For the new terminals of integration I get:

$\displaystyle 1 = tan\theta , \theta = tan^{-1}0 = \frac{\pi}{4}$

$\displaystyle 0 = tan\theta , \theta = tan^{-1}1 = 0$

Subbing into the origanl integral:

$\displaystyle \int_{tan^{-1}1}^{tan^{-1}0}{\frac{1}{sec^2\theta+1}sec^2\theta d\theta}$

$\displaystyle (1)\int_{tan^{-1}1}^{tan^{-1}0}{\frac{sec^2\theta }{sec^2\theta+1}d\theta}$

$\displaystyle \int_{tan^{-1}1}^{tan^{-1}0}{\theta d \theta}$

$\displaystyle \frac{\theta^2}{2}$ from $\displaystyle 0$ to $\displaystyle \frac{\pi}{4}$

Resulting in:

$\displaystyle \frac{(\frac{\pi}{4})^2}{2}-\frac{0}{2} = \frac{\pi^2}{32}$

Checking with my calculator shows that I should have come to:

$\displaystyle \frac{\pi}{4}-tan^{-1}(\frac{1}{2})$

My suspicion is that I mishandled the 1 in the denominator. Am I correct in this thinking?

2. I think that you have gone too far. Simple substitution x'=x+1 will solve the problem in a few moments (clue: arctan(x)).

3. I'm not sure I follow you.

4. let's go over it. x'=x+1. dx'=dx. Your integral becomes integral over (1/(1+x'^2)) when x' is between 1 to 2, which gives you immediately arctan(2)-arctan(1).

I have just assumed that you are familiar with the fact that $\displaystyle \int_{a}^{b}{\frac{1}{1+x^2}dx=atan(b)-atan(a).$

5. Hello, MechEng!

Your substitution and subsequent algebra is off . . .

$\displaystyle \displaystyle{\int_{0}^{1}{\frac{dx}{x^2+2x+2}$

We have: .$\displaystyle \displaystyle{I \;=\; \int^1_0 \frac{dx}{(x+1)^2+1}}$

Let $\displaystyle x+1 \:=\:\tan\theta \quad\Rightarrow\quad dx \:=\:\sec^2\!\theta\,d\theta$

. . and: .$\displaystyle (x+1)^2+1 \:=\:\tan^2\!\theta+1 \:=\:\sec^2\!\theta$

We have: .$\displaystyle \tan\theta \:=\:x+1$

. . When $\displaystyle x = 0\!:\;\tan\theta \:=\:1 \quad\Rightarrow\quad \theta \,=\,\frac{\pi}{4}$

. . When $\displaystyle x = 1\!:\;\tan\theta \:=\:2 \quad\Rightarrow\quad \theta \,=\,\tan^{\text{-}1}\!2$

Substitute: .$\displaystyle \displaystyle{I \;=\;\int^{\tan^{\text{-}1}\!2}_{\frac{\pi}{4}}\frac{\sec^2\!d\theta}{\sec ^2\!\thetta} \;=\;\int^{\tan^{\text{-}1}\!2}_{\frac{\pi}{4}}d\theta }$

. . . . . . . . . . .$\displaystyle =\;\theta\,\bigg]^{\tan^{\text{-}1}\!2}_{\frac{\pi}{4}} \;=\; \tan^{\text{-}1}\!2 - \dfrac{\pi}{4}$