# Integration by Trig Substitution

• Mar 17th 2010, 03:30 AM
p75213
Integration by Trig Substitution
I am trying to integrate the following indefinite integral. 1/(x^2+2x+2)^2. I have this equal to 1/((x+1)^2+1)^2. The sides of the triangle are thus: hypotenuse = (x+1)^2+1,side A = x+1,side B = 1. Therefore with trig substitution I get the following equation to integrate: secant^2 theta/secant^2 theta. However I should get secant^2 theta/secant^4 theta. Can somebody point out where I have screwed up?
• Mar 17th 2010, 03:47 AM
General
Quote:

Originally Posted by p75213
I am trying to integrate the following indefinite integral. 1/(x^2+2x+2)^2. I have this equal to 1/((x+1)^2+1)^2. The sides of the triangle are thus: hypotenuse = (x+1)^2+1,side A = x+1,side B = 1. Therefore with trig substitution I get the following equation to integrate: secant^2 theta/secant^2 theta. However I should get secant^2 theta/secant^4 theta. Can somebody point out where I have screwed up?

• Mar 17th 2010, 04:12 AM
p75213
x = tan theta - 1 -> secant^2 theta dtheta = dx
secant theta = (x+1)^2 + 1
• Mar 17th 2010, 05:11 AM
General
Quote:

Originally Posted by p75213
x = tan theta - 1 -> secant^2 theta dtheta = dx
secant theta = (x+1)^2 + 1

Note that: there is a square on the whole denominator in the original integral ..
• Mar 17th 2010, 10:06 PM
p75213
Yes. The legs of the triangle are wrong. I guess I am asking how to arrive at the correct length of each leg. This one has me stumped. I'm sure it's not difficult but for some reason I just can't see it.