Maths HELP - C4 Integration

I'm really stuck on these questions :/

d through f are just inverse chain rule

g and h are too if you know how to work with odd powers of sin or cos

For I consider the fourth root

Awesome, I'll give that a shot!

If you don't mind me asking, how did you know what to you by looking at it? I mean how did you look at d-f and tell that you need to use the inverse chain rule?

You have g(f(x)) multiplied by f'(x)


The first one does need to be turned into Cos(2x)Sin^-2(2x) first

What do you mean?

The first one does need to be turned into Cos(2x)Sin^-2(2x) first

Why do you need to do that if you can integrat cosec(ax+b)cot(ax+b) = -1/a x cosec(ax + b)

Yup .... I forget that one ... It is so easy to derive using inverse chain rule

Formula sheet rules ok

Ah cool! So I got -1/2cosec(2x) but when I put the limits in I managed to get some long decimals in my answer, am I plugging it into the calculator wrong or?

(In my calculator I did sin(1/2 x pi/3) x -1/2 etc..)

It should be sin(2 x pi/3) then do 1/that

For the second one I got ln(2x-1) but when I substitute the limits in I can't seem to get the right answer! Where am I going wrong?

For the second one I got ln(2x-1) but when I substitute the limits in I can't seem to get the right answer! Where am I going wrong?

What do you get if you differentiate ln(2x-1)

You should see your error

You would get 1/(2x-1) x 2 = 1/4x-2 ... oh

so can you take a factor of 2 out

2ln(2x-1)

You would get 1/(2x-1) x 2 = 1/4x-2 ... oh

so can you take a factor of 2 out

2ln(2x-1)

the question is wrong unless I cannot see well

the integral does not converge between these limits

This is irrelevant in this context (sadly)

At A2 Maths there is no learning related to the validity of an integral .... This is just a question in an exercise designed to test the student's understanding of the inverse chain rule

Hmmmmmmmmm

1/(2x-1) x 2 = 2/(2x-1)

So you need to divide your answer by 2 not multiply it

???

then what answer will you give when there is no answer for these limits...