Integration help needed

[Guest]

I cant integrate that integral
[tex]\int_{sqrt(2)/2}^{1 }\frac{sqrt(1-x^2)}{x^4} dx[/tex]

I think that i must use Trig Substitution to substitute [tex]sqrt(1-x^2)[/tex] . So [tex]sqrt(1-x^2)[/tex] can be written as [tex]sqrt(1^2-sin^2)[/tex] and then i use that if i have [tex]sqrt(a^2-x^2)[/tex] x=asin(u) . And i have [tex]sqrt(1-sin^2)=sqrt(cos^2)[/tex] , but what should i do with the denominator? If x=sin(u) in the denominator i will have
[tex]sin^4(u)[/tex] and in the end i will have [tex]\int_{sqrt(2)/2}^{1 }\frac{sqrt(cos^2)}{sun^4(u)}cos(u)du[/tex]
Becuase x=sin(u) from there i find dx=cos(u)du.So in the numerator :[tex]\int_{sqrt(2)/2}^{1 }\frac{sqrt(cos^2)}{sin^4}[/tex]
But what to do with the denominator? Or i am getting it all wrong?
Some help please, i am struggling. I know that the limits of integration have to be changed too.

[Math Tutor]

Here is a sample solution of the indefinite integral:

2.png (31.09 KiB) Viewed 1310 times

[Guest]

Yes,WolframAlpha,but i dont get why after the substitution he gets [tex]\int_{}^{ }cot^2(u)csc^2(u)du[/tex] Where he gets [tex]u=sin^-1[/tex] . And then he gets that: [tex]\int_{}^{ }cot^2(u)csc^2(u)du[/tex] but i want to know how and why!