1. ## trig substitution

Good afternoon all,

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

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

Through completing the square I have:

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

My chosen substitution is:

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

Using this substitution I get:

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

For the new terminals of integration I get:

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

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

Subbing into the origanl integral:

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

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

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

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

Resulting in:

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

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

$
\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 $\int_{a}^{b}{\frac{1}{1+x^2}dx=atan(b)-atan(a).$

5. Hello, MechEng!

Your substitution and subsequent algebra is off . . .

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

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

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

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

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

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

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

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

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