Integration using method of substitution

[Ry122]

Problem:
(tdt)/(4-t^4)^(1/2)
Attempt:
I want the derivate of whatever i make u equal to, to equal something outside of u therefore I will factorize the denominator to equal -(-2+t^2)(2+t^2) and make u equal to (2+t^2) so that du=2tdt
Balance the equation so that one side is equal to the numerator
(1/2)du=tdt
Now I have
1/2$$\int$$ = 1/(-(-2+t^2)u)^1/2
How do I apply this elementary integral in this situation?:
$$\int$$ 1/(a^2-x^2)^1/2 dx = sin^-1(x/a) + C
I don't know what to do with the u in
1/(-(-2+t^2)u)^1/2

If it helps the final answer is supposed to be:
(1/2)sin^-1(t^2/2)

 

[Dick]

Put u=t^2 first. Then you can apply the arcsin formula.

 

[Ry122]

Could you elaborate? Thanks

 

[Dick]

Could you elaborate? Thanks

Uh, u=t^2. du=2*t*dt. There's your t for the numerator. What does the denominator become?

 

[Ry122]

How can u be made to equal t^2? Can't you only make u equal to a whole polynomial, not just part of it?

 

[Dick]

You can make u whatever you want. What does the denominator become in terms of u?

 

[Ry122]

If u = t^2
then
(-(-2+u)(2+u))^(1/2)?

 

[Dick]

Yes. But no need to factor it. So now you have (1/2)*du/(4-u^2)^(1/2). Now apply the arcsin formula.

 

[Ry122]

Can the arsin formula only be applied if the variable in the denominator has an exponent of 2?

 

[Dick]

Yes. But if you do the substitution u=t^2 then the denominator variable does have an exponent of 2. That was the whole point.

 

[Ry122]

Can you tell me how I would go about solving
(sin(t))^4 x (cos(t))^5) dt
if i make u = sin t
du=cost dt
but that doesn't work because cos(t) has an exponent of 5

 

[Defennder]

$$sin^4t cos t = (sin^2 t)^2 cos t = \frac{1}{2}^2 (1-cos2t)^2 cos t$$

Expand that out. Change the square of the trigo term into a non-square trigo function of a different angle each time you get a sin or cos squared. (What the heck is that trigo identity called?)

Then make use of some trigo identities to express the final integrand into something easily integrable.

 

[Dick]

Use cos^2(t)=(1-sin^2(t)).

 

[cakesama]

double angle formula?

 

[Ry122]

$$sin^4t cos t = (sin^2 t)^2 cos t = \frac{1}{2}^2 (1-cos2t)^2 cos t$$

Expand that out. Change the square of the trigo term into a non-square trigo function of a different angle each time you get a sin or cos squared. (What the heck is that trigo identity called?)

Then make use of some trigo identities to express the final integrand into something easily integrable.

How did you get
$$(cos t)$$

from

$$(cos t)^5$$

 

[Defennder]

Ok, my mistake it should be $$(sintcost)^4 cos t = \frac{1}{2}^4 (sin(2t))^4 cos t = \frac{1}{2}^4 \frac{1}{2}^2(1-cos 4t)^2cos t$$

Just continue from here.

 

[Ry122]

I don't understand how you get that at all. Could you please explain it to me?

 

[Defennder]

Use these formulae:

$$sin(t) cos(t) = \frac{1}{2}sin(2t)$$
$$sin^2(t) = \frac{1}{2}(1-cos(2t))$$