Video Transcript
Determine the indefinite integral of nine sin of 𝑥 over two minus two tan of two 𝑥 plus eight all cubed multiplied by nine over two times the cos of 𝑥 over two minus four sec squared of two 𝑥 with respect to 𝑥.
We can see that our integrand is very complicated in this case. We won’t be able to evaluate this integral directly. In fact, trying to simplify our integrand will not make this integral any easier. We’ll need to use one of our different methods to evaluate this integral. To evaluate this integral, we need to notice that nine over two times the cos of 𝑥 over two minus four sec squared of two 𝑥 is actually the derivative of the inner part of our composite function. It’s the derivative of nine sin 𝑥 over two minus two tan of two 𝑥 plus eight.
Noticing this fact will give us the motivation to try using a 𝑢-substitution to evaluate this integral. We’ll start with 𝑢 is equal to nine sin 𝑥 over two minus two tan of two 𝑥 plus eight. Next, we’ll differentiate both sides of this with respect to 𝑥, using the fact that the derivative of the sin of 𝑎𝑥 for any constant 𝑎 is 𝑎 times the cos of 𝑎𝑥 and the derivative of the tan of 𝑎𝑥 for any constant 𝑎 is 𝑎 times the sec squared of 𝑎𝑥. This gives us d𝑢 by d𝑥 is equal to nine over two times the cos of 𝑥 over two minus four times the sec squared of two 𝑥. And we can see that this is the second factor in our integrand.
Remember that d𝑢 by d𝑥 is not a fraction. However, when we’re using an integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑢 is equal to nine over two cos 𝑥 over two minus four sec squared of two 𝑥 d𝑥. We’re now ready to evaluate our integral by using our 𝑢-substitution. First, the inner component of our first factor is equal to 𝑢. Next, we reasoned that nine over two cos of 𝑥 over two minus four sec squared of two 𝑥 d𝑥 is actually equal to d𝑢.
So combining both of these, our integral just becomes the integral of 𝑢 cubed with respect to 𝑢. And we can actually integrate this by using the power rule for integration, which tells us if 𝑛 is not equal to negative one, the integral of 𝑢 to the 𝑛th power with respect to 𝑢 is equal to 𝑢 to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐶. We add one to our exponent of 𝑢 and divide by this new exponent of 𝑢. In our case, our value of 𝑛 is three. So we get 𝑢 to the fourth power divided by four plus 𝐶.
And we could leave our answer like this. However, remember, our original integral was in terms of 𝑥. So we’ll use our substitution 𝑢 is equal to nine sin 𝑥 over two minus two tan of two 𝑥 plus eight to rewrite our answer in terms of 𝑥. This gives us one-quarter times nine sin of 𝑥 over two minus two tan of two 𝑥 plus eight to the fourth power plus 𝐶. And this is our final answer.
Therefore, by using integration by substitution, we’ve shown the integral of nine sin of 𝑥 over two minus two tan of two 𝑥 plus eight cubed times nine over two cos of 𝑥 over two minus four sec squared of two 𝑥 with respect to 𝑥 is equal to one-quarter times nine sin of 𝑥 over two minus two times the tan of two 𝑥 plus eight to the fourth power plus our constant of integration 𝐶.
