Another simple trigonometric substitution gone wrong

[tangibleLime]

Homework Statement

\int_0^1 \! 7x\sqrt{x^2+4} dx

Homework Equations

The Attempt at a Solution

Noticing that the radical is of the form x^2 + a^2, I know to use a*tan\theta.

x = 2tan\theta

dx = 2sec^2\theta d\theta​

Then I simplified the radical to put it in terms of a trig function.

\sqrt{x^2+4}

\sqrt{4tan^2\theta+4}

I know that this should actually be + 1 instead of - 1, but since it's something squared, I assume it will be positive in the end.

\sqrt{4(tan^2\theta-1}

2sec\theta​

Putting this information back into my original integral and pulling out the 7...

7 \int 2tan\theta*2sec\theta*2sec^2\theta d\theta dx​

Pulling out all of those 2's...

56 \int tan\theta*sec\theta*sec^2\theta d\theta​

In trying to come up with a way to proceed, I figured I'd try a u-substitution. If I substitute u = sec\theta, then du = sec\theta*tan\theta d\theta which would take care of most of the integral.

56 \int u^2 du

56 \frac{u^3}{3}

\frac{56sec^3\theta}{3} + C​

I'm a little bit at a loss about how to continue from here, and it makes me feel like I did something wrong in the steps above. I now have an expression in terms of theta instead of x. I know that sec = 1/cos, and I assume sec^3 = 1/cos^3, but that cube is throwing me off. Any hints would be great, thanks.

 

[Mark44]

Your trig substitution will work, but it's not the best way for this problem. I'll take a closer look in another post, but in the meantime, the best approach for this integral is an ordinary substitution.

u = x² + 4, du = 2x dx