# Another simple trigonometric substitution gone wrong

## Homework Statement

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

## 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.

Last edited:

Mark44
Mentor
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