integration (1 Viewer)

[jkwii]

got a query on which substitution to use and how to do it in this question
2
∫ dx/(x-1)(x^2-2)^0.5
1.5

btw the part with the x's is all the denominator. the only part in the numerator is the dx.

 

[Mark576]

Use u = rt. [x-1]

 

[jkwii]

i know this looks like a simple Q but ∫ dx/(1+tanx) the answer has got logs and lots of crap in it... need help.

 

[Slidey]

Use the substitution u=tan(x/2)
2du=sec^2(x/2) dx
dx=2du/(1+u^2)

2*Int (1-u^2)/[(1+u^2)(1+2u-u^2)] du
-(u^2-2u)+1=-(u-1)^2+2=-(u-1+sqrt(2))(u-1-sqrt(2)), thus:
2*Int (1-u^2)/[(1+u^2)(u-1+sqrt(2))(u-1-sqrt(2)] du
Partial fractions:
(1-u^2)/[(1+u^2)(u-1+sqrt(2))(u-1-sqrt(2)]
=
A/(u-1-sqrt(2)) + B/(u-1+sqrt(2)) + (Cu+D)/(1+u^2)

You know what? I think I'll let you solve that. It is solvable. Even if you end up with E/(1+u^2) - use the substitution u=tan(v) to get Int Ecos^2(v) (and remember the double-angle formula).

I looks like you'll to get some mix of the cosine of inverse tan + the log of tangents, and remember that even once you get an answer in u, you still have to sub in u=tan(@/2).

If they ask this in an exam, I'll eat my keyboard.

 

[jkwii]

the answer was like 0.5 ln (x^2-2x-1) - 0.5 ln (x^2 +1) + 0.5 x + C

WTF!!!

 

[malady]

From Int 2(1-u^2)/(1+u^2)(1-u^2+2u), by partial fractions u have Int[(u-1)/(u^2-2u-1)-(u-1)/u^2+1) so have 0.5 ln(u^2-2u-1)-0.5 ln(u^2+1) + arctan(u) and then put back u= tan(x/2)

 

[Slidey]

Lol. I was tempted to try that but figured it wouldn't work. Well help any.

Funnily enough, that makes this question easier than if it were simply Int dx/(cos(x)+sin(x)). I.e. 2*Int du/(1+2u-u^2)

 

[jkwii]

thanks it helped a lot