I hate Trigonometric Subsitution

[Tom McCurdy]

How would you do

$$\int (1-x^2)^{3/2}$$


after you get to the point where you have made the triangle and you have

$$x=sin\theta$$

and

$$dx=cos\theta$$

 

[dextercioby]

Yeah,Tom,with the substitution it becomes
$$\int \cos^{4}x dx$$

What would u do next??

Daniel.

 

[Tom McCurdy]

thats the problem... lol
I was able to get to that point... I just don't know how to get to the answer... I checked it on my calcutor and found a really long answer.

 

[Pyrrhus]

Tom, how about

$$\int \cos^{4}x dx$$

$$\int (\cos^{2}x)^{2} dx$$

$$\int ( \frac{1 + \cos 2x}{2})^{2} dx$$

Edit: It was redudant, This is better, Yes dextercioby, i fixed it now, sorry for that :tongue2:

 

[dextercioby]

Call your integral 'I'.I'm sure u still remember that:
$$\cos^{4}x=\cos^{2}(1-\sin^{2}x)$$
Then 'I' is split into 2 integrals:
$$I=\int (\cos^{2}x -\cos^{2}x\sin^{2}x) dx$$
Call the first integral by $I_{1}$
$$I_{1}=\int \cos^{2} x dx$$

HINT:$$\cos^{2}x=\frac{1}{2}(1+\cos 2x)$$

Call the second integral by $I_{2}$
$$I_{2}=\int \cos^{2}x\sin^{2}x dx$$

HINT:$$\sin 2x=2\sin x\cos x$$
$$\sin^{2} 2x=\frac{1}{2}(1-\cos 4x)$$

Then
$$I=I_{1}-I_{2}$$

To return to the initial variable (remember u made the substitution x->sin x),simply make the 'x' in your result for 'I' go to $\arcsin x$ (i.e.reverse the substitution).

Good luck!I'm going to go to bed now.By the time i get up,u'd better have it done...

Daniel.

EDIT:Cyclovenom,don't give him bad advice... :grumpy:

 

[Tom McCurdy]

...

sorry didn't finish it... or the rest of the homework

I got Iintegral one down to

$$1/2 \int (x + \frac{sin2x}{2} )$$

I started

I2 but I wasn't sure where the $$sin^22x$$ hint came in since it was just sin

Anyway I prob just didn't spend ennuogh time on it... I willl actually work on it some before that class...

 

[dextercioby]

Okay,let's check it:
$$I_{1}=\frac{1}{2}\int (1+\cos 2x) dx=\frac{1}{2}(x+\frac{1}{2}\sin 2x)=\frac{x}{2}+\frac{\sin 2x}{4}$$(1)

Yous still has the integral sign.Maybe it was a typo...

$$I_{2}=\int \sin^{2}x\cos^{2}x dx=\frac{1}{4}\int \sin^{2} 2x =\frac{1}{8}\int (1-\cos 4x) dx=\frac{1}{8}(x-\frac{1}{4}\sin 4x)=\frac{x}{8}-\frac{\sin 4x}{32}$$ (2)

Now 'I' is the difference between the first intergral and the second,viz.
$$I=(\frac{x}{2}-\frac{x}{8})+\frac{\sin 2x}{4}+\frac{\sin 4x}{32} =\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}$$ (3)

I'll let u invert the substitution $x\rightarrow \arcsin x$.

Daniel.