Determine the integral of negative eight cos of four 𝑥 all over two sin of four 𝑥 minus five with respect to 𝑥.
We’re asked to evaluate the integral of the quotient of two trigonometric functions, and we don’t know how to do this in the general case. So, there’s a few different methods we could try. For example, we could try using trigonometric identities to rewrite our expression. We might also be tempted to use integration by substitution. However, in this case there is actually a fast simpler method we could notice.
Consider what would happen if we differentiate our denominator with respect to 𝑥. We know for any real constant 𝑎, the derivative of the sin of 𝑎𝑥 with respect to 𝑥 is equal to 𝑎 times the cos of 𝑎𝑥. So, differentiating our denominator with respect to 𝑥, we can set our value of 𝑎 equal to four and multiply this by two. This gives us eight cos of four 𝑥. And we can see this is a scale factor of our numerator. In fact, it’s our numerator multiplied by negative one. And we know an integral rule to help us evaluate integrals of this form.
We know the integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑥 plus the constant of integration 𝐶. We’re now almost ready to use this to evaluate our integral. We’ll set 𝑓 of 𝑥 to be the function in our denominator. That’s two sin of four 𝑥 minus five. Then, 𝑓 prime of 𝑥 is eight cos of four 𝑥, which appears in our numerator.
However, this isn’t exactly in the form of our integral rule. We still have a factor of negative one inside of our integral. However, we know we can just take this outside of our integral. Now, we can see this is negative one times the integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥. So, we can just apply our integral rule.
So, by setting 𝑓 of 𝑥 and using our integral rule and multiplying through by negative one and remembering we can add our constant of integration 𝐶, this gives us negative the natural logarithm of the absolute value of two times the sin of four 𝑥 minus five plus our constant of integration 𝐶.