Can someone explain this question about trig substitution?

I am not sure what I am doing wrong!

Here is the question:

Evaluate the definite integral

$$\int_0^{7/2}\sqrt{49-x^2} \ dx.$$

What I have gotten to is that the integral (through trig substitution) is $$7\int \cos(x) \ dx = 7[\sin x].$$

Where I am having trouble is finding the limits that go with the new integral. I have the answer, it is $$\frac{49 \pi}{12} + \frac{49}{8}(3)^{1/2}.$$ I would like to understand how to figure this question out. Thanks!

Here is the work I've done so far You did the substitution $x =7\sin \theta$. But do not forget to convert the $dx$ part!

To figure the bounds out, wonder for which values of $\theta$ you have $$7\sin \theta = \frac72\\ 7\sin \theta = 0$$and the simplest solution to this question is $$\theta = \frac \pi 2 \\ \theta = 0$$ which are the bounds of the new integral (in the same order).

• You mean $7\sin\theta=\frac{7}{2}$ in the first equation, right? – SBareS Dec 11 '14 at 19:58
• indeed! thanks :) – mookid Dec 11 '14 at 20:02

You probably did the substitution $x = 7 \sin(t)$, for $\displaystyle 0 \leq t \leq \frac{\pi}{2}$. To find the inferior and superior limits all you have to do is substitute $x$ for the respective inferior and superior limits and see what $t$ turns out to be.

If $x=0$ then $\sin(t) = 0$ and $t=0$. If $\displaystyle x = \frac{7}{2}$ then $\displaystyle \sin(t) = \frac{1}{2}$ and $\displaystyle t = \frac{\pi}{6}$.

Don't forget that you need to deal with the $dx$ in your original integral when you do a trig. substitution. It looks like the substitution you made is to set $x = 7 \sin (\theta)$. Then

$\frac{dx}{d\theta} = 7\cos(\theta)$.

Therefore,

$dx = 7\cos(\theta) \, d\theta$.

Next, we have to deal with the limits of integration. First, plug $x=0$ into $x = 7 \sin (\theta)$. This gives $0 = 7\sin\theta$, so $0 = \sin\theta$ and $\theta = 0$. Next, plug $x = 7/2$ into $x = 7 \sin (\theta)$. This gives $7/2 = 7\sin\theta$, so $1/2 = \sin\theta$, and $\theta = \pi / 6$. Using these substitutions in the original integral, we get:

$\int_0^{7/2} \sqrt{49-x^2} \,dx = \int_{\theta = 0}^{\theta = \pi / 6} \sqrt{49-(7\sin\theta)^2} (7 \cos\theta) \,\, d\theta = \int_{\theta = 0}^{\theta = \pi / 6} 49\cos^2 \theta \,\, d\theta$.

If you evaluate this integral you'll get the correct answer. If you need assistance evaluating the integral feel free to leave a comment below.

• Ok, I see what I did! Thank you so much! – Kris Dec 11 '14 at 20:10
• You're welcome, happy to help! – Gecko Dec 11 '14 at 20:56