Integration by Substitution

[Idealistic]

Integrate
x(1 - x[sup:3b7kotfv]4[/sup:3b7kotfv])[sup:3b7kotfv]1/2[/sup:3b7kotfv]dx

I've tried u = 1 - x[sup:3b7kotfv]4[/sup:3b7kotfv] but all that got me was

1/4du = x[sup:3b7kotfv]3[/sup:3b7kotfv]dx

x = (1 - u)[sup:3b7kotfv]1/4[/sup:3b7kotfv]

1/4du = (1 - u)[sup:3b7kotfv]3/4[/sup:3b7kotfv]dx

1/4du/(1 - u)[sup:3b7kotfv]3/4[/sup:3b7kotfv] = dx

hence

u[sup:3b7kotfv]1/2[/sup:3b7kotfv](du)/(1 - u)[sup:3b7kotfv]3/4[/sup:3b7kotfv] - Can't do much with this

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[galactus]

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

Try the sub \(\displaystyle u=x^{2}, \;\ \frac{du}{2}=xdx\)

 

[Idealistic]

Thanks

 

[Idealistic]

I got (1 - u[sup:1r6k3gtp]2[/sup:1r6k3gtp])[sup:1r6k3gtp]1/2[/sup:1r6k3gtp]du, is this the antiderivative of 1/(Arcsinu)?

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[galactus]

No, it is not. It does not work that way. Even though \(\displaystyle \int\frac{1}{\sqrt{1-u^{2}}}du=sin^{-1}(u); \;\ \;\ \int\sqrt{1-u^{2}}du\neq \frac{1}{sin^{-1}(u)}\)

Try the trig sub \(\displaystyle u=sin(t), \;\ du=cos(t)dt\)

\(\displaystyle \frac{1}{2}\int\sqrt{1-u^{2}}du\)

\(\displaystyle \frac{1}{2}\int\sqrt{1-sin^{2}(t)}\cdot cos(t)dt\)

\(\displaystyle \frac{1}{2}\int cos^{2}(t)dt\)

Now, integrate and resub.

To integrate \(\displaystyle cos^{2}(t)\) use the identity \(\displaystyle cos^{2}(t)=\frac{cos(2t)+1}{2}\)

 

[galactus]

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

Come to think of it, here is another way.

This method gets rid of the radical, but I don't see where it's really that much easier. Perhaps.

Let \(\displaystyle x=(1+u^{2})^{\frac{-1}{4}}, \;\ dx=\frac{-u}{2(u^{2}+1)^{\frac{5}{4}}}du\)

When we make the subs and expand, we get:

\(\displaystyle \frac{1}{2}\int\frac{1}{(u^{2}+1)^{2}}du-\frac{1}{2}\int\frac{1}{u^{2}+1}du\)

Now, integrate away. Maybe you will like this a little better than the other way.

 

[Idealistic]

No, it is not. It does not work that way. Even though \(\displaystyle \int\frac{1}{\sqrt{1-u^{2}}}du=sin^{-1}(u); \;\ \;\ \int\sqrt{1-u^{2}}du\neq \frac{1}{sin^{-1}(u)}\)

Try the trig sub \(\displaystyle u=sin(t), \;\ du=cos(t)dt\)

\(\displaystyle \frac{1}{2}\int\sqrt{1-u^{2}}du\)

\(\displaystyle \frac{1}{2}\int\sqrt{1-sin^{2}(t)}\cdot cos(t)dt\)

\(\displaystyle \frac{1}{2}\int cos^{2}(t)dt\)

Now, integrate and resub.

To integrate \(\displaystyle cos^{2}(t)\) use the identity \(\displaystyle cos^{2}(t)=\frac{cos(2t)+1}{2}\)

I understand this method a little bit clearer, I just never would have thought to substitute sint for u, after substituting u for x? I didn't know it was mathematically kosher. Hence first year student...

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[mmm4444bot]

… never would have thought to substitute sin(t) for u, after substituting u for x? I didn't know it was mathematically kosher …

As long as a symbol is not already "in use", we are free to replace ANY expression with ANY such symbol at ANY time.

 

[Idealistic]

The only thing is that the question, which I would have never thought to be so complicated, is actually an indefinate integral (from 0 to 1), so I finished subbing (left off from (1 - u[sup:32260i5h]2[/sup:32260i5h])[sup:32260i5h]1/2[/sup:32260i5h], then letting u equal sin(t), then getting cos[sup:32260i5h]2[/sup:32260i5h](t), and using the trig identity) and got this:

1/2(1 + cos(2t)) and the antiderivative to be:

1/2(t - sin(2t)/2) and then to make life easier I did this: sin(2t) = 2sin(t)cos(t)

1/2(t - sin(t)cos(t)) and subbed "u" back in:

-1/2(t - udu) and then x back in:

1/2(t - (x[sup:32260i5h]2[/sup:32260i5h])2x) **My concern is that I have x and t in the same function!? If I want to complete the definite integral I have to do F(b) - F(a) but my F's are in terms of two different variables.

 

[galactus]

There should not be two variables. You can just change limits of integration accordingly and then when you evaluate

\(\displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos^{2}(t)dt\) you have your solution.

 

[Idealistic]

There should not be two variables. You can just change limits of integration accordingly and then when you evaluate

\(\displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos^{2}(t)dt\) you have your solution.

but I didn't think the area under cos[sup:2uvl2xx7]2[/sup:2uvl2xx7](t) from 0 to pi/2 was equal to the area under x(1 - x[sup:2uvl2xx7]4[/sup:2uvl2xx7])[sup:2uvl2xx7]1/2[/sup:2uvl2xx7] from 0 to 1. I graphed both functions and the area under their curves seem quite different, cos[sup:2uvl2xx7]2[/sup:2uvl2xx7](t) had larger area under its curve

***Oh wait, its one half cos[sup:2uvl2xx7]2[/sup:2uvl2xx7](t)


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