Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    A bit of help would be appreciated for the following:

    1) Show that INT1 and 0) x/(1 + x)^1/2 = (2/3)(2 - root2)

    2) INT: sec^(2)x/(1 + tan x)^3

    3) INT: (sin x + 2 cos x)^2

    For number three, the answer is (5/2)x + (3/4)sin2x - cos 2x. I get the same answer, apart from the - cos 2x bit. I multiplied out the bracket, so the 2nd term should be 4 sin x cos x. Then when I integrated this, I get either 4 sin^(2)x or 4 cos^(2)x. Could somewhere tell me how I correctly integrate this please?

    Cheers everyone
    Offline

    0
    ReputationRep:
    (Original post by foowise)
    A bit of help would be appreciated for the following:

    1) Show that INT1 and 0) x/(1 + x)^1/2 = (2/3)(2 - root2)

    2) INT: sec^(2)x/(1 + tan x)^3

    3) INT: (sin x + 2 cos x)^2

    For number three, the answer is (5/2)x + (3/4)sin2x - cos 2x. I get the same answer, apart from the - cos 2x bit. I multiplied out the bracket, so the 2nd term should be 4 sin x cos x. Then when I integrated this, I get either 4 sin^(2)x or 4 cos^(2)x. Could somewhere tell me how I correctly integrate this please?

    Cheers everyone
    For the last bit you were asking about, change it into a 2sin2x. And integrate that, to -cos2x
    Offline

    2
    ReputationRep:
    (Original post by foowise)
    A bit of help would be appreciated for the following:

    1) Show that INT1 and 0) x/(1 + x)^1/2 = (2/3)(2 - root2)
    Use the substitution u²= 1+x
    2u.du/dx = 1
    dx = 2u.du

    New limits: x=0, u=1...x=1, u=√2

    =INT (√2 and 1) (u²-1 / u) . 2udu

    =2 [INT (u²-1) du]

    = 2 [u³/3 - u] (limits: √2 and 1)

    = 2 [(2√2)/3 - √2] - [(1/3) - 1]

    simplifies to (2/3)(2-√2)
    • Thread Starter
    Offline

    0
    ReputationRep:
    Cheers guys, any chance of help for number 2?

    And, just one more thing, could someone show that 4 cos 2x/sin^(2)2x = cosec^(2)x - sec^(2)x. I'm getting better at all those proofs, but this one has still got me. Cheers
    Offline

    1
    ReputationRep:
    (Original post by foowise)
    Cheers guys, any chance of help for number 2?

    And, just one more thing, could someone show that 4 cos 2x/sin^(2)2x = cosec^(2)x - sec^(2)x. I'm getting better at all those proofs, but this one has still got me. Cheers
    2) ∫(sec^2(x))/((1+tanx)^3) dx

    = ∫(1+tanx)'*(1+tanx)^3 dx

    =((1+tanx)^4)/4 + k
    Offline

    1
    ReputationRep:
    (Original post by foowise)
    Cheers guys, any chance of help for number 2?

    And, just one more thing, could someone show that 4 cos 2x/sin^(2)2x = cosec^(2)x - sec^(2)x. I'm getting better at all those proofs, but this one has still got me. Cheers
    cosec²(x) - sec²(x) = (cos²x-sin²x)/(sin²x*cos²x)

    =(cos(2x))/(0.25*(2sinxcosx)²)

    =4*cos2x/(sin2x)²

    quod erat demonstrandum
    Offline

    2
    ReputationRep:
    (Original post by foowise)
    Cheers guys, any chance of help for number 2?

    And, just one more thing, could someone show that 4 cos 2x/sin^(2)2x = cosec^(2)x - sec^(2)x. I'm getting better at all those proofs, but this one has still got me. Cheers
    L.H.S: = 4 cos 2x/sin^(2)2x

    = [4 (cosx)^2 - (sinx)^2] / 4(sinx)^2(cosx)^2

    = (cosx)^2 - (sinx)^2 / (sinx)^2(cosx)^2

    = [(cosx)^2 / (sinx)^2(cosx)^2] - [(sinx)^2 / (sinx)^2(cosx)^2] <-----split up the fraction

    = [1/(sinx)^2] - [1/(cosx)^2]

    = (cosecx)^2 - (secx)^2
 
 
 
Poll
Do you think parents should charge rent?

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

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