Year 13 Maths Help Thread

I'm not sure how to solve these integration questions.
This is some of my working out. Thanks.

For 7), I think you're getting mixed up with $\sin^{-1} x$ and $\left(\sin x\right)^{-1}$.

$\sin^{-1} x$ is "inverse sine" and you can't differentiate it using the method you tried.

Instead, you have $x=\sin u$ so find $\frac{dx}{du}$. There's no need to rearrange to get $u$ in terms of $x$ like you did.


For 8), $\frac{du}{dx} = e^x = u$

So try using $u$ instead of $e^x$ in your working.

For 7), I think you're getting mixed up with $\sin^{-1} x$ and $\left(\sin x\right)^{-1}$.

$\sin^{-1} x$ is "inverse sine" and you can't differentiate it using the method you tried.

Instead, you have $x=\sin u$ so find $\frac{dx}{du}$. There's no need to rearrange to get $u$ in terms of $x$ like you did.


For 8), $\frac{du}{dx} = e^x = u$

So try using $u$ instead of $e^x$ in your working.

Sorry I'm back home now. For 7) I've tried this so far.

You absolutely cannot take cos u through the integral sign like that. It is not a constant! Integrate by parts.

Edit: Oh and you didn't change the limits either.

Oh thank you. How would I integrate inverse sine by parts?

I was suggesting you integrate $\int u \cos u \, du$ by parts.

Ok thanks.

Don't forget to change the limits of integration first.

I tried to integrate by parts and I got usinu + cosu which doesn't seem right. Could you show me the steps please?

That is correct. What makes you think it doesn't seem right?

When I tried to substitute u=sin^(-1)x back into the answer, it didn't make sense

You don't need to substitute anything back in. Just change the limits from x = something to u = something.

Oh ok so do I substitute 1 and 0 into x=sinu?

Substitute those numbers into $u=\sin^{-1} x$ . Make sure you are in radians mode.

Its been a while since ive done this sort of stuff (simplifying fractions)...... and I keep getting it wrong now (ikr)

Can someone tell me what im doing wrong please?

I multiplied the RHS fraction by x-1 then added both:
4 + 2(x-1)
-------------
(x+1)(x-1)

Maybe im being stupid but i dont think that simplifies to 2 / (x+1)

Factorise the numerator and you will find something cancels out.

4 + 2(x-1) 2(2+(x-1)) 4
------------- = ------------------- = ------- ????
(x+1)(x-1) (x+1)(x-1) (x+1)