This A-level maths is funny compared to IB HL...

Right so you got 1/(cost)^3 dx ... you also know that dx=cost dt
giving 1/cos²t dt

you also need to find the new bounds since x=sint
giving lower bound 0 and upper bound pi/4

So youre left to find the definite integral between 0 and pi/4 of 1/cos²t dt...

Which integrates to tant. now you plug in pi/4 and 0. and subtract both results.

tan(pi/4)-tan(0) = 1-0 = 1

If you don't understand a step, feel free to quote and ask.

(edited 11 years ago)

Reply 8

davros

16

Original post by IShouldBeRevising_

Can some one please post a full solution so I can work through it...

Why don't you follow the suggestions given and post your working and then we can comment.

You were nearly there at the start, except you put in sin x where you should have put

sin \theta

. You can then simplify the denominator, and you also need to convert dx into

d\theta

using

dx/d\theta

Reply 9

giobru

Original post by Giveme45

This A-level maths is funny compared to IB HL...

Right so you got 1/(cost)^3 dx ... you also know that dx=cost dt
giving 1/cos²t dt

you also need to find the new bounds since x=sint
giving lower bound 0 and upper bound pi/4

So youre left to find the definite integral between 0 and pi/4 of 1/cos²t dt...

Which integrates to tant. now you plug in pi/4 and 0. and subtract both results.

tan(pi/4)-tan(0) = 1-0 = 1

If you don't understand a step, feel free to quote and ask.