7-56. Trigonometric substitutions Evaluate the following integ...

Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

30. ∫ x³√(1 - x²) dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral by using trigonometric substitution. We are given the integral of X cubed multiplied by the square root of 4 minus X squared. Now the square root term is in the form of a squared constant minus X squared, and that means that we can define the trigonometric substitution to be X's equal to a multiplied by sine theta. Here, A is going to be the squared constant that we are subtracting from X2. Now, in this case here, that constant is 4. This tells us that A2 is equal to 4, and that means that A is going to equal to 2. So, we can redefine our trigonometric substitution to be X is equal to 2 multiplied by sin of theta. And we need to come up with a substitution of DX in terms of theta. We will take the derivative of both sides of this equation with respect to X, and we will be left with DX is equal to 2 multiplied by cosine of theta D theta. So if we were to apply our trigonometric substitution to the integral, we can rewrite the integral to be X cubed, which is going to be 2 sin theta cubed, which further simplifies to 8 sine cubed theta. Multiplied by the square root of 4 minus X2, which is going to be 4 squared theta, multiplied by the substitution of DX, which is going to be 2, cosine of theta D theta. So there are a few things that we can do to further simplify this integral. First, for the square root, we can factor out of 4 and simplify it using the square root. We can rewrite the square root to be 2, multiplied by the square root of 1 minus sine square theta, and by the Pythagorean identity, we can further rewrite this to be 2 multiplied by the square root of cosine square theta, which will further simplify to be 2 cosine of theta. We are going to have 3 coefficients inside the integrand in total. We have 2 multiplied by 2, multiplied by 8, and that is going to give us a total product of 32. So, we can further simplify this integral to be 32 multiplied by the integral of sine cubed theta. Multiplied by cosine of theta, multiplied by cosine of theta D theta, and multiplying the cosine terms together is going to give us cosine squared. So we will have an integrand of sine cubed theta multiplied by cosine squared theta D theta. Now we will approach this using our trig trigonometric methods. So, sinecube theta can be rewritten as a product of signs. We can rewrite the integran to be 32, multiplied by the integral of sine squared theta. Multiplied by cosine square theta. Multiplied by sin of theta D theta. By using trigonometric identities, we can rewrite sine square theta to be 1 minus cosine squared theta. So, we will have 32 multiplied by the integral of 1 minus cosine squared theta, multiplied by cosine squared theta. Multiplied by sin of theta D theta. In order to simplify this integral, we will use a second substitution. Let's go ahead and allow W to equal to cosine of theta. Then DW will equal to negative sin of theta d theta, which can also be rewritten as negative DW is equal to sin of theta D theta, which we have inside the integrand. So by bringing a negative to the front of the integral and applying our W substitutions, we can rewrite this integral to be -32 multiplied by the integral of 1 minus W2 multiplied by W2DW. We will go ahead and distribute W2d into the integrand, and that is going to give us -32 multiplied by the integral of W2 minus W to the fourth power DW. And by taking the antiderivative of each term within the integral, we will be left with -32 multiplied by the quantity, 13. Multiplied by W cubed minus 15th. Multiplied by W to the 5th power, and because we are evaluating a definite indefinite integral, we will add a general constant C. Now, we are going to distribute negative or what we will do first is we will go ahead and rewrite W back into terms of you. Or back in terms of theta. So we allow W to equal to cosine theta, and that is going to give us -32 multiplied by 1/3, multiplied by cosine cubed theta minus 1/5 multiplied by cosine to the 5th theta plus C. Next, what we will go ahead and do is we will go ahead and rewrite cosine theta back in terms of X. Now, earlier, we said that our trigonometric substitution is going to be X is equal to 2 sin of theta. If we solve for sin of theta in that substitution, we will have sin of theta is equal to X divided by 2. Now we can make a right triangle with respect to sin of theta. Recall that sin of theta is the opposite side divided by hypotenuse. So if we create a right triangle with respect to these values. This tells us that the opposite side to theta has a value of X, and the hypotenuse has a value of 2. And by using the Pythagorean identity, we get that the missing side is going to be the square root of 4 minus X squared. Now, cosine theta, with respect to this right triangle, is going to be the adjacent side of the divided by the hypotenuse. That is going to be the square root of 4 minus X2 divided by 2. And if we were to resubstitute this back into our integral or back into our solution, we'll be left with -32. Multiplied by 1/3, multiplied by the square root of 4 minus x 2 divided by 2 raise to the third power minus 1/5 multiplied by the square root of 4 minus X2 divided by 2 raise to the 5th power plus C. And finally, if we were to distribute and simplify the coefficients, the simplest form of our solution is going to be -1 divided by 15, multiplied by the quantity 4 minus X2d raised to the three halves power. Multiplied by X + 3 X 2 + C. This is going to be the simplest solution to the given indefinite integral. So I hope this video helps you in understanding how to use trigonometric substitution, and we will go ahead and see you all in the next video.

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
30. ∫ x³√(1 - x²) dx

Key Concepts

Trigonometric Substitution

Integral of a Polynomial

Pythagorean Identity

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