Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    0
    ReputationRep:


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    Offline

    3
    ReputationRep:
    (Original post by IShouldBeRevising_)


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    dx/d(theta) = ?
    Then multiply by this.
    • Political Ambassador
    Offline

    3
    ReputationRep:
    (Original post by IShouldBeRevising_)


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    Hint:

    x=\sin \theta \Rightarrow \dfrac{dx}{d\theta} = \cos \theta \Rightarrow dx = \cos \theta \ d\theta

    You don't just substitute sin x .

    Note that you also have to change the limits.
    Offline

    3
    ReputationRep:
    (Original post by IShouldBeRevising_)


    10i

    I have subbed in sin x

    And got 1/cos^3x ... What next?
    Before the 'next':
    After substitution you should to get \frac{1}{\cos^3 \theta} for the integrandus.

    And what will you get for the integrating factor of dx with substitution
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can some one please post a full solution so I can work through it...
    Offline

    0
    ReputationRep:
    Im stuck myself I got to this 1/(cos^2t)^3/2dx
    1/(cos^2t)^3/2 (1/cost) dx
    Offline

    15
    ReputationRep:
    (Original post by IShouldBeRevising_)
    Can some one please post a full solution so I can work through it...
    Try what people have suggested first. Cannot give full solutions here.
    Offline

    0
    ReputationRep:
    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.
    • Study Helper
    Offline

    15
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    (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.


    wrong, the limits you have are incorrect remember that x=sin theta so theta = arcsin x and hence you get the limits
    Offline

    0
    ReputationRep:
    (Original post by giobru)
    wrong, the limits you have are incorrect remember that x=sin theta so theta = arcsin x and hence you get the limits
    True, admitedly made a stupid mistake there, didn't pay enough attention, gj for pointing out
 
 
 
Poll
Which pet is the best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.