Integral with trig-substitution

• Aug 20th 2012, 12:48 PM
gralla55
Integral with trig-substitution
So I'm pretty much stuck with an integral problem. It was on the form (1 + (2t)^2)^0,5, using trig substitution I got to the integral 1/2(sec x)^3. Continuing with integration by parts, I finally got the solution (INT)1/2(sec x)^3 = 1/4(sec x)(tan x) + 1/4 ln |sec x + tan x| which should be correct. However, when I try to substitute back, using t = 1/2 tan x, my answer does not equal the answer in my book. And I can't seem to figure out what I did wrong... I've uploaded a picture of the last part of the problem, with the answer in my book below my answer.

Any help would be greatly appreciated!
• Aug 20th 2012, 02:03 PM
Soroban
Re: Integral with trig-substitution
Hello, gralla55!

Quote:

$\int \sqrt{1 + (2t)^2}\,dt$

I would avoid those silly fractions . . .

$\text{Let }2t \:=\:\tan\theta \quad\Rightarrow\quad t \:=\:\tfrac{1}{2}\tan\theta \quad\Rightarrow\quad dt \:=\:\tfrac{1}{2}\sec^2\!\theta\,d\theta$
And: . $\sqrt{1+(2t)^2} \:=\:\sqrt{1+\tan^2\!\theta} \:=\:\sqrt{\sec^2\!\theta} \:=\:\sec\theta$

Substitute: . $\int \sec\theta\left(\tfrac{1}{2}\sec^2\!\theta\,d \theta \right) \;=\;\tfrac{1}{2}\int\sec^3\!\theta\,d\theta$

. . . . . . . . . $=\;\tfrac{1}{4}\bigg(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg) + C$ .[1]

Back-substitute: . $\tan\theta \,=\,\frac{2t}{1} \,=\,\frac{opp}{adj}$

$\theta$ is in a right triangle with: . $opp = 2t,\;adj = 1$
Hence: . $hyp \,=\,\sqrt{1+4t^2}$
And: . $\sec\theta \,=\,\frac{hyp}{adj} \,=\,\frac{\sqrt{1+4t^2}}{1} \,=\,\sqrt{1+4t^2}$

Substitute into [1]: . $\tfrac{1}{4}\bigg(\sqrt{1+4t^2}\cdot 2t + \ln\left|\sqrt{1+4t^2} + 2t\right|\bigg) + C$

. . . . . . . . . . . . . $=\;\tfrac{1}{2}\:\!t\sqrt{1+4t^2} + \tfrac{1}{4}\ln\left|2t + \sqrt{1+4t^2}\right| + C$
• Aug 20th 2012, 02:14 PM
gralla55
Re: Integral with trig-substitution
Hi and thank you for very much for your help! I see you got the exact same answer I got, except I factored in the (1/2) under the first square root sign. What I can't seem to understand, is how the book gets from ln(2t + SQRT(1 + 4t^2)) to ln(t + SQRT((1/4 + t^2)). How are these two things the same??

Thanks again!
• Aug 20th 2012, 03:23 PM
skeeter
Re: Integral with trig-substitution
they're not "the same" ...

$\ln(2t + \sqrt{1 + 4t^2}) =$

$\ln \left[2t + \sqrt{4\left(\frac{1}{4} + t^2\right)}\right] =$

$\ln \left[2t + 2\sqrt{\frac{1}{4} + t^2}\right] =$

$\ln \left[2\left(t + \sqrt{\frac{1}{4} + t^2}\right)\right] =$

$\ln{2} + \ln\left(t + \sqrt{\frac{1}{4} + t^2}\right)$

... but they do differ by a constant.
• Aug 20th 2012, 03:42 PM
gralla55
Re: Integral with trig-substitution
How did I not think of that.... Thank you very much! Now I can finally go to sleep, haha.