Integration Help: Tips for Solving Integration Questions | 6 Examples

[Zoe-b]

Homework Statement

I have various integration questions to do, and I do not want answers at all, just pointers on where to start.. I have 6 ones, and have only got anywhere with 2 of them (nt the two below)
Find integral of with respect to x:
1.
((x-1)/(x+1))^1/2 * 1/x

2.
1/(2 - sin²x)(2 + sinx - sin²x)

Homework Equations

integration by substitution?

The Attempt at a Solution

For the first I tried u = x +1 but this doesn't go anywhere- I wouldn't know how to integrate it even without the extra 1/x term so really don't know where to start.

For the second I tried the substitution u = sin x thinking maybe this would lead to something where I could use partial fractions (it doesn't), and then there's no cos x to even allow the substitution in the first place.

My integration is REALLY rusty, so although I can manage integration by parts and simpler stuff, I seem to have completely lost the technique of what to substitute where.


One of the other ones is to find the integral of x * arcsin x dx. For this I get:
1/2(x² * arcsinx + x (1-x²)^1/2 - arcsin x ) + C

I can't find a problem in the working (I used integration by parts where u = x, u' = 1, v' = arcsin x, v = x * arcsin x + (1-x²)^1/2), but the maths program I use occasionally to check things came up with different coefficients in the answer.

 

[gabbagabbahey]

For the first I tried u = x +1 but this doesn't go anywhere- I wouldn't know how to integrate it even without the extra 1/x term so really don't know where to start.

Try the substitution u=\sqrt{x+1} instead and show your work.

For the second I tried the substitution u = sin x thinking maybe this would lead to something where I could use partial fractions (it doesn't), and then there's no cos x to even allow the substitution in the first place.

It's not clear exactly what the 2nd one is supposed to be...do you mean \int\frac{2+\sin x-\sin^2x}{2-\sin^2 x}dx or \int\frac{dx}{(2-\sin^2 x)(2+\sin x-\sin^2 x)} ?

In any case, there is a well-known trig identity that relates \cos x to \sin x, so there is nothing to stop you from making your substitution.

One of the other ones is to find the integral of x * arcsin x dx. For this I get:
1/2(x² * arcsinx + x (1-x²)^1/2 - arcsin x ) + C

I can't find a problem in the working (I used integration by parts where u = x, u' = 1, v' = arcsin x, v = x * arcsin x + (1-x²)^1/2), but the maths program I use occasionally to check things came up with different coefficients in the answer.

Post your workings then, because your final result is incorrect.

 

[hunt_mat]

I think that for the trigonometric integral, you can use partial fractions.