Today, we continue to look at integrals using the different integration techniques shown in Calculus 2. In this problem, we will find the \(\int_{0}^{8} \sqrt{64-x^{2}}dx\). In order to evaluate this integral, we will need to use a trig substitution, then we will need to use reduction formulas.

Hopefully you will join me each day as I go through these integrals in both a post and in a video. Along the way, I hope you learn something, and, if you are taking Calculus 2, you are in a great position to do very well on any upcoming exams you may have. You can find more Calculus 2 study help posts here. I also have a decision tree available so that you can follow along as we determine what technique of integration to use. I also have a useful list of formula sheet available.

**Find \(\int_{0}^{8} \sqrt{64-x^{2}}dx\).**As we start this problem, we want to make sure that we can still use the fundamental theorem in order to evaluate the definite integral by finding an anti-derivative. That is, we need to check that the integrand \(\sqrt{64-x^{2}}\) is defined on interval \([0,8]\). In this case, we note that \(64-x^{2} \geq 64-8^{2}=0\) for all \(x\) between \(0\) and \(8\), so we can take the square root. Therefore, we can use the fundamental theorem.

In order to find the anti-derivative of \(\sqrt{64-x^{2}}\), we again go through our decision tree.

- Can you guess the answer?

We will work under the assumption that we cannot guess the answer. - Is there an inside function?

Here there is an inside function, namely \(i=64-x^{2}\).

We will therefore attempt the substitution using \(i=64-x^{2}\). This gives us \(di=-2x dx\), so \(dx=\frac{di}{-2x}\). We then get

\begin{align*}

\int \sqrt{64-x^{2}}dx=\int \sqrt{i}\frac{di}{-2x}.

\end{align*}

In this case, we are not able to cancel the \(x\)s out. Furthermore, the remaining \(x\)s do not look like the \(i\) we defined, so we will need to start over making different choices.

**A different kind of substitution**

We begin the process of going through our decision tree again.

- Can you guess the answer?

We will work under the assumption that we cannot guess the answer. - Is there an inside function?

We saw last time that using substitution here won’t work, so we move onto the next question. - Is there a product of function?

No there isn’t. - Is this a rational function?

No it isn’t. - Is there something of the form \(a^{2}+1\)?

No there isn’t. - Is there something of the form \(1-a^{2}\)?

While \(64-x^{2}\) doesn’t exactly look like this, we can transform it into this form.

We will therefore simplify the equation so that we can use the substitution \(a=\sin(\theta)\). To do this, we find

\begin{align*}

\sqrt{64-x^{2}}&=\sqrt{64(1-\frac{1}{64}x^{2})} \\

&=\sqrt{64}\sqrt{1-(\frac{1}{8}x)^{2}}.

\end{align*}

Therefore, we let \(\frac{1}{8}x=\sin(\theta)\). This gives us \(x=8\sin(\theta)\) and \(dx=8\cos(\theta)d(\theta)\). We then get

\begin{align*}

\int \sqrt{64-x^{2}}dx&=\int 8 \sqrt{1-(\sin(\theta))^{2}}8\cos(\theta)d(\theta) \\

&=64 \int \sqrt{\cos^{2}(\theta)}\cos(\theta)d(\theta) \\

&=64 \int \cos^{2}(\theta)d(\theta).

\end{align*}

Now that we have simplified, we see that there are no more \(x\)s in the integral, so we do have a new integral in terms of \(\theta\). Furthermore, the new integral looks simpler than the original, so we will start over and find this new integral.

**Find \(\int \cos^{2}(\theta)d(\theta)\).**

We now start our decision tree again and find

- Can you guess the answer?

We will work under the assumption that we cannot guess the answer. - Is there an inside function?

While there is an inside function of \(\cos(\theta)\), we note that the using this will not allow us to cancel out the \(x\)s. - Is there a product of function?

Yes, there is. - Is this the product of trig function?

Yes, we have the product of cosines.

We will therefore attempt to use the reduction formulas. In this case, we will need the formula

\begin{align*}

\int \cos^{n}(x) dx=\frac{\cos^{n-1}(x)\sin(x)+(n-1)\int \cos^{n-2}(x)dx}{n}.

\end{align*}

Using this formula, we find that

\begin{align*}

\int \cos^{2}(\theta)dx&=\frac{\cos(\theta)\sin(\theta)+\int 1d\theta}{2} \\

&=\frac{\cos(\theta)\sin(\theta)+\theta}{2}+c.

\end{align*}

Therefore, we have found the integral of \(\cos^{2}(\theta)\).

**Combining this together.**

We can now combine all the information we have together to get our original integral. The first thing we need to notice is that \(\theta\) was a variable that we introduced into the problem, and we will need to get rid of it. To do this, we recall \(\frac{1}{8}x=\sin(\theta)\), so we get \(\theta=\sin^{-1}(\frac{1}{8}x)\). We then get that

\begin{align*}

\int \sqrt{64-x^{2}}dx&=\frac{\cos(\theta)\sin(\theta)+\theta}{2}+c \\

&=\frac{\cos(\sin^{-1}(\frac{1}{8}x))\sin(\sin^{-1}(\frac{1}{8}x))+\sin^{-1}(\frac{1}{8}x)}{2}+c \\

&=\frac{\cos(\sin^{-1}(\frac{1}{8}x))(\frac{1}{8}x)+\sin^{-1}(\frac{1}{8}x)}{2}+c.

\end{align*}

Then, we find that

\begin{align*}

\int \sqrt{64-x^{2}}dx&=64(\frac{\cos(\sin^{-1}(\frac{1}{8}x))(\frac{1}{8}x)+\sin^{-1}(\frac{1}{8}x)}{2}|_{0}^{8} )\\

&=64(\frac{\cos(\sin^{-1}(1))(1)+\sin^{-1}(1)}{2}-\frac{\cos(\sin^{-1}(0))(0)+\sin^{-1}(0)}{2})\\

&=64(\frac{\cos(\frac{\pi}{2})1+\frac{\pi}{2}}{2}-\frac{\cos(0)*0+0}{2} )\\

&=16\pi.

\end{align*}

We have now found that the definite integral, or the area under, \(\sqrt{64-x^{2}}\) on the interval \([0,8]\) is exactly \(\frac{\pi}{4}\).

**Conclusion**

In this problem, we noted that we did have to try a few integration techniques before we finally realized that we would have to simplify the problem using a trig substitution before integrating. We then have to find the integral of \(\cos^{2}(x)\) by using reduction formulas. We, therefore, had to use multiple techniques in order to find anti-derivative. When we we done with this, we still had to use this to find the definite integral by evaluating at the limits of integration.

I hope you learned something from this post and found it helpful. If you did, make sure to share it and the video with anyone else you know that may need some extra help with integrals.