Integration by subs. (1 Viewer)

[mchew92]

I need help understanding this question
I have the teachers solution but i dont understand the actual integrating part

Int (cos x sin x) / (2-sin(^2) x)^-0.5 dx

ie. i dont get the part where you: [ Int u(2-u^2)^-0.5 du ]

please explain. thx in advance

 

[oly1991]

i think this is it:
let u=sinx , therefore du=cosx.dx

(int) u/(2-u^2)^-0.5
= (int) u(2-u^2)^-1.5
= 2u(2-u^2)^-0.5 + C
=2u/(2-u^2)^0.5 + C
= 2sinx/(2-sin(^2)x)^0.5 + C

 

[mchew92]

The answer is -(2-sin(^2) x)^0.5 + C
i think you did something wrong

 

[study-freak]

I need help understanding this question
I have the teachers solution but i dont understand the actual integrating part

Int (cos x sin x) / (2-sin(^2) x)^-0.5 dx

ie. i dont get the part where you: [ Int u(2-u^2)^-0.5 du ]

please explain. thx in advance

Sure that u got the highlighted part right?

Coz the whole expression then = Int (cosxsinx)(2-sin^(2)x)^0.5 dx

 

[mchew92]

oh sorry i think i left out substitution by letting u=sinx

 

[scardizzle]

upon subbing u = sinx you get u/(2-u^2)^1/2

then you use the reverse chain rule e.g. (intergral) f'(x). (f(x))^n = (f(x))^n+1

therefore you get -(2-u^2)^1/2

if you have trouble understanding differentiate (2-u^2)^1/2 you'll end up the the question but with a negative sign

sub sinx back in

EDIT: is your first name michael by any chance?

 

[Drongoski]

Is the integral (1) or (2) ??

 

[study-freak]

oh sorry i think i left out substitution by letting u=sinx

No, I was asking a simple question like
are you sure the question says a/b^-1, which is just ab anyway? coz it's a strange way of writing
and it has to be a/b, not a/b^-1 looking at the answer

I.e. what drongoski is asking

 

[mchew92]

Its number (1) yeah my bad
Yes i am scardizzle, who are u?

 

[scardizzle]

Hey Michael, it's Jeffrey how's the study going?

 

[addikaye03]

Can anyone think of a way to integrate this without sub, i just thought of one.

 

[cutemouse]

t-results =D

Would be quite messy though.

 

[Drongoski]

 

[scardizzle]

possibly a bunch of trig identities might simplify to a nicer term?