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Question Video: Finding the Integration of a Function Involving Trigonometric Functions Using Integration by Substitution Mathematics • Higher Education

Determine ∫((6 sin 2π‘₯ βˆ’ 2 tan 2π‘₯)⁡ (12 cos 2π‘₯ βˆ’ 4 secΒ² 2π‘₯)) dπ‘₯.

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Video Transcript

Determine the indefinite integral of six times the sin of two π‘₯ minus two times the tan of two π‘₯ raised to the fifth power multiplied by 12 cos of two π‘₯ minus four sec squared of two π‘₯ with respect to π‘₯.

The question gives us a very complicated-looking integral. We might be tempted to try distributing our exponents to simplify our integrand. However, simplifying our integrand in this case would be very complicated. We have a binomial expression raised to the fifth power multiplied by another binomial. So we want to look for a different method which is easier. And to find this easier method, we need to notice something about our integrand. We need to notice our second factor of 12 times the cos of two π‘₯ minus four times the sec squared of two π‘₯ is actually the derivative of the inner part of our composite function. It’s the derivative of six sin of two π‘₯ minus two tan of two π‘₯.

This gives us motivation to try integrating this by substitution. We’ll set 𝑒 to be six sin of two π‘₯ minus two tan of two π‘₯. Next, we’ll differentiate both sides of this expression with respect to π‘₯ by using our trigonometric derivative rules. For a constant π‘Ž, the derivative of the sin of π‘Žπ‘₯ is equal to π‘Ž times the cos of π‘Žπ‘₯. And for a constant π‘Ž, the derivative of the tan of π‘Žπ‘₯ is equal to π‘Ž times the sec squared of π‘Žπ‘₯. This gives us d𝑒 by dπ‘₯ is equal to 12 times the cos of two π‘₯ minus four times the sec squared of two π‘₯.

Now, remember, d𝑒 by dπ‘₯ is not a fraction. However, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials. d𝑒 is equal to 12 times the cos of two π‘₯ minus four sec squared of two π‘₯ dπ‘₯. And we can then see that this appears in our integral. This means we’re now ready to evaluate our integral by using integration by substitution. First, we say 𝑒 to be the inner part of our composite function. And this then told us that 12 cos of two π‘₯ minus four sec squared of two π‘₯ dπ‘₯ is actually equivalent to d𝑒. So by using our 𝑒 substitution, we can rewrite our integral as the integral of 𝑒 to the fifth power with respect to 𝑒.

And we can evaluate this integral by using the power rule for integration, which tells us if 𝑛 is not equal to negative one, the integral of 𝑒 to the 𝑛th power with respect to 𝑒 is equal to 𝑒 to the power 𝑛 plus one divided by 𝑛 plus one plus our constant of integration 𝐢. We add one to our exponent of 𝑒 and then divide by this new exponent. In our case, our exponent of 𝑒 is five. So we get one over six multiplied by 𝑒 to the sixth power plus 𝐢. Remember though, our original integral was in terms of π‘₯. So we want to write our answer in terms of π‘₯. We’ll do this by using our substitution 𝑒 is equal to six sin of two π‘₯ minus two tan of two π‘₯. And this gives us one over six times six sin of two π‘₯ minus two tan of two π‘₯ to the sixth power plus 𝐢. And this is our final answer.

Therefore, we’ve shown, by using integration by substitution, the integral of six times the sin of two π‘₯ minus two tan of two π‘₯ to the fifth power multiplied by 12 times the cos of two π‘₯ minus four sec squared of two π‘₯ with respect to π‘₯. Is equal to one over six times six sin of two π‘₯ minus two times the tan of two π‘₯ raised to the sixth power plus a constant of integration 𝐢.

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