Problem 40 Medium Difficulty

One CAS produces $-\frac{2}{15} \tan x-\frac{1}{15} \sec ^{2} x \tan x+\frac{1}{5} \sec ^{4} x \tan x$
as an antiderivative in example $3.7 .$ Find $c$ such that this equals our antiderivative of $\frac{1}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+c$.

Answer

$
-\frac{2}{15} \tan x-\frac{1}{15} \sec ^{2} x \tan x+\frac{1}{5} \sec ^{4} x \tan x-\frac{1}{3} \tan ^{3} x-\frac{1}{5} \tan ^{5} x
$

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Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendental Functions

Chapter 6

Integration Techniques

Section 3

Trigonometric Techniques of Integration

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Video Transcript

so I want to show that these two expressions are the same for a particular value of C. So start with the left side and the idea is we can notice that the left side has seeking functions involved. So using the dragon identity, namely, the one that says that one plus Tansi in squared of X equal secrets quarterbacks, we will be able to write the left side in terms of just the tangent function. So starting with the left side, we can see that we have negative to 15th times tangent of X minus 1 15 times seeking squared of X But keep in mind that seeking squared of X was one plus tangent square to vex And then times the Tanja next I'm gonna bring the tangent X to be next to the negative 1/15 And then the sequence squared of X will become the binomial one plus tangent squared of X and that will help our distribution In a moment when we write that expression that we have plus 1/5 again I'm gonna bring the tangent X next to the 1/5 and then for seek it to the fourth ex Keep in mind that that would be equivalent to seek in squared of X squared. So we're going to replace the seek and squared events with one plus tangent square of X, and now square that expression. So now we're going to expand their expression and simplify. So I have negative to 15th Time's tangent events. Next, we're going to distribute negative 50 negative 1/15 Tanja necks teach term and the one plus tangent squared of X to give us negative 1/15 tangent of X minus 1 15 times Tangent, cubed of X because when we multiply like basis, we at their exponents and before multiplying the 1/5 tangent X through the next parentheses. Keep in mind that we have to square that parentheses first, which means this where the first term in the second term. But we'll also have a metal term that we need to think about so have one plus tangent Square debts times one plus tangent Squared X. Let me move that back up. So again, the one plus tangent squared X Quantity Square. Remember that that means to take one plus tangent squared X and multiply it by one plus tangent squared X so As a result, we will obtain one plus two times tangent square necks, less tangent to the fourth of X. So now we'll be able to distribute that peace through, so we now have negative to 15th Time's Tangent of X minus 1/15 Time's Tangent of Eggs minus 1 15 times Tangent, Cuba Events plus 1/5 Tangent of X would distribute to the first term. Then we have plus 2/5 attention chilled of X. My distribute to the second term and then distributing to 1/3 term will have positive 1/5 tangent to the fifth of X. Next, we want to come by light terms, so start by noticing the tangent X terms. So there are three tangent. Next terms that will combine together, which means to actor coefficients were combined. Their coefficients, which will require common denominator, says, Keep in mind that 1/5 is equivalent to 3/15 so I have a negative to 15th a negative 1/15 and a positive 3/15 Tanja necks, which will combine 20 Next, we can look at our tangent cubed terms again. Combining their coefficients needed a common denominator, keeping in mind that truth. If 2/5 would be equivalent. Excuse me to 6 15 6 15 and a negative 1/15 would give us a positive 5 15 tangent of X, which will reduce further in a moment. And then we still have the 1/5. So a positive 1/5 tangent, 1/5 of X Or, in other words, we have zero plus 1/3 tangent CUC of X plus 1/5 Tangent, 1/5 of X, which looks like our right side of our equation as long as our see value is zero.

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