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Evaluate the integral. $ \displaystyle \int_0^…

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Problem 29 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int x \sqrt{1 - x^4}\ dx $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

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Video Transcript

Let's evaluate the integral of X times the square root of one minus X to the fourth. Before we go to a tricks up, we should try to use up first. Let's take you to be X squared so that do you is two x dx equivalently do you over to as xd X and we see those terms in the original inte grand. Here is X and DX. So we can write this as one half inaugural Radical one minus u squared, do you? And then now for this new integral, we can go ahead and use the tricks up. So here we should take you to be one signed data, then do U S coast in data. So plugging this in we have one half integral one minus sine square times coastline Taito So far, next step using Pythagorean theorem one minus sine squared is coastline squared and we know that the square root of cosine squared. It's just co sign. So we have one half coastline square data. So now we have a trigger inaugural cosine squared. So for this it will be best to use the identity. One plus co signed up to data all divided by two having no identity running out of room. Here, let me go to the next page. Simplifying. We have 1/4 one political sine two theta. Now we're ready to integrate and the role of one is just data and the integral of cosine two theta. It's signed to data over to, but we also have a four here. So let it go. But say eight pussy now we'll eventually draw the triangle to get back into the variable X. But even with the triangle will have a problem here because we have signed to data. So in order to rewrite this in terms of X, we can use the half angle or the double angle formula to sign data cosign data and then two divided by eight, just 1/4. Now we can draw the triangle. So recall that are tricks up was the equal sign data. If we want, we can also write this as you over one. Uh huh. So we have you for the opposite one for the hypotenuse. And by Pythagorean theorem, we could find a adjacent side age. We know that a squared plus U squared is one so h equals square root one minus u squared. So coming back to our original problem. So we have 1/4 data we don't really get from the triangle. We can get that from our tricks up. Solve this for Veda by taking sine inverse on both sides. It's data equals some members of you. And then we have signed data, which is you and then co sign data, which is h divided by one. So it's just a judge, and then the last step is to rewrite you in terms of X. So I'm running out of room again. Let me go to the next page. Yeah, recall our u substitution from the very beginning. So plugging this in for you and our previous expression we have 1/4 sine inverse of you becomes sine inverse of x squared. And then we have you which becomes X squared or excuse me? We have X squared, which becomes messing up here. We have previously had you and the numerator that becomes X squared. And then we had square root of one minus u squared that becomes one minus X to the fourth. So u squared equals X squared squared. Excellent fourth and we were dividing by four plus e. And there we go. This is our answer

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Problem

Evaluate the integral. $ \displaystyle \int_0^…

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Related Courses

Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 7

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Kayleah Tsai

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Kayleah Tsai

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Heather Zimmers

Kayleah Tsai

Joseph Lentino

James Stewart

Calculus: Early Transcendentals