I was browsing math videos on You Tube and found this nicely done presentation on doing integration by parts twice. Check it out.

From: Dave McMahon
1:31 PM

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for some functions f and g. For example:

In this case we can identify:

How do you compute the derivative in this case? The chain rule tells us that we compute the derivative in the following way:

Let’s compute some examples. Consider:

Here we have:

And

Hence:

Here’s another example. Suppose you’re given:

In this case we have:

Meanwhile:

Putting everything together we get:

Let’s do one more. It’s basically the same as the previous example with a minor twist:

Proceeding as before we get:

Meanwhile

And so the final result is:

Implicit differentiation is a technique used to compute a derivative when a function y = f(x) is given indirectly by an equation that relates x and y. For example suppose that you are given:

The equation must be solved for y to obtain an explicit solution, and in this case there are two solutions. The procedure called implicit differentiation works by differentiating both sides of the equation and then solving for dy/dx, so an explicit expression of the derivative is not given. In the video some examples illustrate this procedure.

From here we can write down their anti-derivatives:

The minus sign comes about because we integrated the derivative of cos x to get this result. That is

but this was equal to minus sin x. The other antiderivative is

Where trig integrals get complicated is when they include products or powers of trig functions. But once again, due to the relationship among derivatives of trig functions, integrating their products is usually fairly easy and only looks complicated at first glance. Let’s consider an example:

This is actually a simple u substitution integral. Let’s take:

Hence the integral becomes:

Replacing u we find that:

The cyclic nature of the derivatives of trig functions also plays a role when doing integration by parts with trig integrals. See the entry on integration by parts with trig functions for an example. In that case you will notice that by parts integration has to be applied twice in order to get the final answer.

Trig integrals involving ratios of trig functions are no different than those involving products. The simplest example is the integral of the tangent function, which can be approached using u substitution. First we write the tangent as the ratio of sin to cosine:

Now we take:

This yields:

Hence:

Now we can use the properties of the natural logarithm to write this in a form usually seen in integral tables, which will get rid of the minus sign for us:

And of course this is just the secant function:

We’ll see in a moment that this integral evaluates to an inverse tangent. We’re going to use two steps to convert this into the following elementary form:

The first step is to bring the denominator into a more appropriate form. We want something like:

where a and b are constants. When you see a problem with a quadratic function in the denominator, try completing the square. In this case we can do the following:

Now we can write the integral as follows:

Now we can to the integral by u substitution. To get it in the form we want, where we can get the form for the inverse tangent integral, we will have to pick u such that we can pull out all the constants in the denominator. This can be done if we take:

We do this because we could transform the integral as follows:

modulo a proportionality factor. Taking the square root of our chosen u substitution, we get:

That is:

Now we can get the result we seek by using this substitution in the original integral. This gives:

Now replace u to get the final result:

If you just set x = 2, you get 0/0. However this limit is done easily by factoring the numerator. Notice that:

We can use this to cancel the troublesome term in the denominator. So we get:

Now we can just set x = 2 and write down the answer:

Sometimes solving calculus limits is a little more tricky. These cases are where, unfortunately, your knowledge of algebra comes in handy. Let’s try:

The approach to calculating calculus limits of this type is really the same-factor the numerator and denominator when possible. Let’s look at the denominator first, with an odd power it’s a little more troublesome. But we can write it like this:

Now we can work on the numerator to see how we can get rid of the x – 1 term. Notice that:

So we have:

Now we just cancel the x – 1 terms and do the limit:

Its always a good idea to start out writing the integration by parts formula:

For now, we can simply ignore the limits of integration. First we take

Then we have

And so

Now we’ve got to apply integration by parts on the integral we have at the end. Let’s go through it in case you aren’t familiar with it. Take

Hence

Now we can put this together with our previous result, and we’ve found that

If we were doing an indefinite integral, we would add a constant of integration to the end and we’d be done. For integration by parts with limits, all we do now is evaluate this expression at the limits of integration. The upper limit was x = 3, which gives:

The lower limit was x =2, which gives

So the answer is found by subtracting the lower limit from the upper limit, and you can verify that

So, to do integration by parts with limits, simply do the integral the way you would in the indefinite case, ignoring the limits of integration. Then when you have your answer at the end, evaluate at each limit and subtract.

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