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

Determine β« ((β24π₯Β³ + 30 sin 6π₯)(β6π₯β΄ β 5 cos 6π₯)β΅) dπ₯.

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### 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. Weβll let π’ be the inner function in our composite function, and then we use the general result for the derivative of cos ππ₯.

And we see that dπ’ by dπ₯, the derivative of π’ with respect to π₯, is negative 24π₯ cubed plus 30 sin six π₯. Remember, dπ’ 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 dπ’ 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 dπ’. And we replace negative six π₯ to the fourth power minus five cos six π₯ with π’.

And we see that our integral becomes really nice. Itβs the integral of π’ to the fifth power dπ’. Well, the antiderivative of π’ to the fifth power is π’ to the sixth power over six. So, the integral of π’ to the fifth power dπ’ is π’ to the sixth power over six plus the constant of integration π. Remember though, our integralβs in terms of π₯, so we replace π’ with negative six π₯ to the fourth power minus five cos of six π₯.

And weβve evaluated our integral. Itβs a sixth of negative six π₯ to the fourth power minus five cos of six π₯ to the sixth power plus π.