Week 4 Discussion Notes

Solution.

First, we want to get rid of the $2$ in the sine term so that all the angles match up. If you remember your trig identities,

Let $u = \ln\p{\tan{x^2}}$, which gives

Our new bounds will be

Now we can do our substitution:

Solution.

This is actually a trig problem in disguise. It becomes a very easy problem if you remember that

You can either draw a picture or use the angle-sum identity to prove it. So, the integral becomes

Solution.

After a glance, this is of the form $\infty \cdot 0$. So, we can rewrite it and apply L'Hôpital's:

Solution.

The limit has the form $\frac{0}{0}$, so we can apply L'Hôpital's straight away:

You might be tempted to use L'Hôpital's again, but if you look ahead, you should see that you basically get your original limit back, so it won't be of any help. This means we have to try to calculate the limit as is. Like one of the problems before, we want to get rid of the $2x$ is the cosine in the numerator. If you recall, $\cos{2x} = 2\cos^2{x} - 1$, so the numerator becomes $2\cos{x} - 4\cos^2{x} + 2$. We're going to want to factor this, so if we let $u = \cos{x}$,

So, the limit becomes

The $2\cos{x} + 1$ term goes to $3$, so it's harmless here, which means we want to turn our attention to the other terms:

Finally, we can calculate our limit:

Solution.

This is of the form $\infty^0$, so let $y = \displaystyle \p{\csc{x}}^{\p{\cos^{-1}{x}}/\ln{x}}$, which gives

The $\cos^{-1}{x}$ term is harmless since it goes to $\frac{\pi}{2}$, so we need to look at the other terms.

So, this means

Going back to our original limit, continuity of $e^x$ gives

Solution.

The limit has the indeterminate form $\frac{\infty}{\infty}$, so we can use L'Hôpital's right away.

Solution.

This is of the form $1^\infty$, so like before, we're gonna set $y = \p{\frac{\sin{x}}{x}}^{1/x^2}$. Then

When $x \to 0^+$, we get the indeterminate form $\frac{0}{0}$, so by L'Hôpital's

Finally, going back to the original limit, we get

Solution.

First, we're going to rewrite the limit as

We know that $\lim_{x\to0} \frac{\sin{x}}{x} = 1$, so we only have to worry about the other factor. It has the form $\frac{0}{0}$, so we can use L'Hôpital's:

So, we get

Solution.

Even though there's only one term, we can still use integration by parts because $\cos^{-1}{x} = \cos^{-1}{x} \cdot 1$. Choose $\diff{v} = \cos^{-1}{x} \,\diff{x}$ isn't going to do anything for us because we don't know how to integrate it (yet), so we need to let $u = \cos^{-1}{x}$ and $\diff{v} = \diff{x}$ instead. Then

So if we integrate by parts, we get

(To calculate the integral, you can use $u = 1 - x^2$.)

Solution.

I want to differentiate $\ln{x}$ because it gives me $\frac{1}{x}$, which plays nicely with all the other terms. So, we want to choose $u = \ln{x}$ and $\diff{v} = \frac{1}{x^3} \,\diff{x}$, which gives

Then the integral becomes

Solution.

This is one of those problems that requires multiple uses of integration by parts and a trick. Here, your choice of $u$ and $\diff{v}$ don't really matter, so I'm going to set $u = e^{2x}$ and $\diff{v} = \sin{x} \,\diff{x}$, which gives

After the first application of integration by parts, we get

We're gonna do the same thing, except with $\diff{v} = \cos{x} \,\diff{x}$ this time, so we get

Notice that we get the same integral on both sides, so we can just solve for it. If we add $4\int e^{2x} \sin{x} \,\diff{x}$ to both sides, we get

Solution.

This problem can be made a lot easier if you remember your trig identities:

so the integral becomes

When doing integration by parts, we don't want to set $\diff{v} = x \,\diff{x}$ because we'll end up with something like $x^2 \cos{2x}$, which is harder. Instead, we'll set $u = x$ and $\diff{v} = \sin{2x} \,\diff{x}$, which gives

So, the integral becomes