Arcsin Integral Format: Solving ∫1/(x^4)(sqrt(a^(2) - x^(2)) for Homework

[Justabeginner]

Homework Statement

Evaluate the integral ∫1/(x^4)(sqrt(a^(2) - x^(2))

Homework Equations

u= dx/sqrt(a^2-x^2)
du= arcsin(x/a)dx

The Attempt at a Solution

Am I even doing this right? And if so, how would I express 1/x^4 in terms of u? Have I taken the right u-substitution? Thanks.

 

[Simon Bridge]

Lets make sure I'm reading that right:

Evaluate the integral ∫1/(x^4)(sqrt(a^(2) - x^(2))
[...]
u= dx/sqrt(a^2-x^2)
du= arcsin(x/a)dx

You are asked to evaluate:
$$\int \frac{dx}{x^4\sqrt{a^2-x^2}}$$ ... and you are using the substitution:
$$u=\frac{dx}{\sqrt{a^2-x^2}}\\ du = \arcsin(x/a)dx$$ ... I have no idea why you would choose such a substitution or how you arrived at it.

The usual way to get rid of a square root would be to use a trig substitution like $x=a\sin\theta$ so that $dx = a\cos\theta\; d\theta$

 

[Justabeginner]

I chose this substitution because one of the formulas I learned is that the integral of dx/sqrt(a^2-x^2) is arcsin(x/a) dx; but it's supposed to be the derivative. My mistake :/

Using trig substitution, I get:

int [1/(x^4)(sqrt(a^2-sin^2(theta))]

I'm not sure how to continue at all.

 

[Simon Bridge]

int [1/(x^4)(sqrt(a^2-sin^2(theta))] would be if you put $x=\sin\theta$ ... that's the wrong substitution.

You can't do just any old substitution.

Here, what you want to do is exploit that $\sin^2\theta + \cos^2\theta = 1$.

Don't forget that the substitution applies to every instance of x and to dx also.