Trig substitution or a different procedure?

**zabooma87** · March 26th 2009, 03:40 PM

Solve:
∫√1+(2x-4)^2dx
Hello everyone, I'm new to this so please bare with me. I'm not sure how to do this problem. I've thought maybe trig substitution but I've been unsuccessful in evaluating the problem. Now, I'm not asking for an answer, but any little bit of guidance/advice would be appreciated. Thanks everyone.

-John

**mollymcf2009** · March 26th 2009, 06:23 PM

Originally Posted by zabooma87

Solve:
∫√1+(2x-4)^2dx
Hello everyone, I'm new to this so please bare with me. I'm not sure how to do this problem. I've thought maybe trig substitution but I've been unsuccessful in evaluating the problem. Now, I'm not asking for an answer, but any little bit of guidance/advice would be appreciated. Thanks everyone.

-John

Hi John! Welcome to the forum!

I'll get you started and see if you can finish.

$\int \sqrt{1 + (2x - 4)^2} dx$

u = 2x-4
du = 2 dx

$\frac{1}{2} \int \sqrt{u^2 + 1} du$

Notice that this is a trig substitution $x = a tan(\theta)$

$\frac{1}{2} \int \sqrt{tan^2(\theta) + 1} \cdot sec^2(\theta) d\theta$

See what you can do from here! Good luck!

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