Thread: Trigonometric Substitution Calc. II Help Please!

integral 8/((4x^2+1)^2) dx

I know you have to let x=1/2tanu
therefore dx = 1/2 sec^2 u du

But I get a wrong answer! Somewhere along the way I get messed up.

Also:

integral e^t/(sqrt(e^(2t)+9)) dt from 0 to ln4

Use a substitution like u=e^t to simplify the problem, then let u = 3tant
and so du = 3sec^2t

I also get a wrong answer for this one. Be careful with the limits.

Originally Posted by joepinsley

integral 8/((4x^2+1)^2) dx

I know you have to let x=1/2tanu
therefore dx = 1/2 sec^2 u du

$\displaystyle \int \frac{4\sec^2{u}}{(\tan^2{u}+1)^2} \, du$

$\displaystyle \int \frac{4\sec^2{u}}{\sec^4{u}} \, du$

$\displaystyle \int 4\cos^2{u} \, du$

$\displaystyle 2\int 1 + \cos(2u) \, du$

$\displaystyle 2u + \sin(2u) + C$

$\displaystyle u = \arctan(2x)$ ...

$\displaystyle 2\arctan(2x) + \sin[2 \arctan(2x)] + C$

$\displaystyle 2\arctan(2x) + 2\sin[\arctan(2x)]\cos[\arctan(2x)] + C$

$\displaystyle 2\arctan(2x) + 2 \cdot \frac{2x}{\sqrt{4x^2+1}} \cdot \frac{1}{\sqrt{4x^2+1}} + C$

$\displaystyle 2\arctan(2x) + \frac{4x}{4x^2+1} + C$