Video: Finding the Integration of a Function Involving Trigonometric Functions Using Integration by Substitution

Determine ∫ ((βˆ’24π‘₯Β³ + 30 sin 6π‘₯)(βˆ’6π‘₯⁴ βˆ’ 5 cos 6π‘₯)⁡) dπ‘₯.

02:00

Video Transcript

Determine the integral of negative 24π‘₯ cubed plus 30 sin six π‘₯ times negative six π‘₯ to the fourth power minus five cos of six π‘₯ to the fifth power with respect to π‘₯.

To evaluate this integral, we need to spot that negative 24π‘₯ cubed plus 30 sin six π‘₯ is the derivative of the inner part of this composite function negative six π‘₯ to the fourth power minus five cos six π‘₯. This tells us we can use integration by substitution to evaluate this integral. We’ll let 𝑒 be the inner function in our composite function, and then we use the general result for the derivative of cos π‘Žπ‘₯.

And we see that d𝑒 by dπ‘₯, the derivative of 𝑒 with respect to π‘₯, is negative 24π‘₯ cubed plus 30 sin six π‘₯. Remember, d𝑒 by dπ‘₯ is not a fraction, but we do treat it a little like one when performing integration by substitution. And we see that this is equivalent to saying that d𝑒 is equal to negative 24π‘₯ cubed plus 30 sin six π‘₯ dπ‘₯. We, therefore, replace negative 24π‘₯ cubed plus 30 sin six π‘₯dπ‘₯ with d𝑒. And we replace negative six π‘₯ to the fourth power minus five cos six π‘₯ with 𝑒.

And we see that our integral becomes really nice. It’s the integral of 𝑒 to the fifth power d𝑒. Well, the antiderivative of 𝑒 to the fifth power is 𝑒 to the sixth power over six. So, the integral of 𝑒 to the fifth power d𝑒 is 𝑒 to the sixth power over six plus the constant of integration 𝑐. Remember though, our integral’s in terms of π‘₯, so we replace 𝑒 with negative six π‘₯ to the fourth power minus five cos of six π‘₯.

And we’ve evaluated our integral. It’s a sixth of negative six π‘₯ to the fourth power minus five cos of six π‘₯ to the sixth power plus 𝑐.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.