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    Hi Im stuck on a few questions now as i approach the end of the chapter can someone guide me through these:

    1 . Find integral in terms of x:

    ∫ x/root(x+1) dx u^2 = x + 1

    2. Use the sub u = x - 1 ∫ x/root(x-1) dx

    3. Use the sub u= sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)
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    (Original post by phiyahless)
    Hi Im stuck on a few questions now as i approach the end of the chapter can someone guide me through these:

    1 . Find integral in terms of x:

    2. Use the sub u = x - 1 ∫ x/root(x-1) dx

    3. Use the sub u= sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)

    1 . Find integral in terms of x:

    ∫ x/root(x+1) dx u^2 = x + 1

    rearrange your substitution to give x=u^2 - 1 (put in top)
    square root your sub. and let the denominator be u
    rearrange your substition in terms of u (x=u^1-1) and differentiage with respect to u>>>dx = 2u du
    sub all back into integral to get integral u^2 -1 / u * 2u du

    integrating gives 2u^3 / 3 -2u
    resubbing u gives...2(x+1)^3 - 2(x+1) + C

    2. Use the sub u = x - 1 ∫ x/root(x-1) dx
    same method as for (1)

    numerator = u+1
    denominator = u^0.5 and dx = 1.du
    resubbing gives integ u/u^0.5 du = integ u^0.5 du = 2u^1.5 /3
    resub u gives integ = 2/3 * (x-1) ^1.5

    3. Use the sub u= sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)

    du / dx = cos x > dx = du / cos x.

    resub into original integral and the cos x on the bottom cancels with the one ni the integral.

    now use identity for cos 2x to get 1-2sin^2 x. you know sin x = u> sub this in:

    integral = u-2/3 u ^ 3 between limits

    BUT...you got to change the limits to u, or change the equation into x from and put the angle limits in...for convenience i will convery it to U limits.

    at pi / 3 u = sin pi / 3 = (rt 3) / 2 and at pi/6 u = sin pi / 6 = 0.5

    put these limits into the integral worked out above...to get:

    -((5+2 (rt2) )/12)

    answers may be wrong... i'm really tired...awwww.

    goodnight people!
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    (Original post by phiyahless)
    Find integral in terms of x:

    1. ∫ x/root(x+1) dx u^2 = x + 1
    2. Use the sub u = x - 1 ∫ x/root(x-1) dx
    3. Use the sub u = sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)
    1.) u^2 = x + 1 --> x = u^2 - 1 --> dx/du = 2u --> dx = 2u du
    ∫ x/root(x+1) dx
    = ∫ (u^2 - 1)/u . 2u du
    = ∫ 2(u^2 - 1) du
    = ∫ 2u^2 - 2 du
    = (2u^3)/3 - 2u + k
    = [2(x + 1)^(3/2)]/3 - 2Sqrt(x + 1) + k

    2.) u = x - 1 --> du/dx = 1 --> dx = du
    ∫ x/root(x-1) dx
    = ∫ (u + 1)/(Sqrtu) du
    = ∫ u^(1/2) + u^(-1/2) du
    = [2u^(3/2)]/3 + 2u^(1/2) + k
    = [2(x - 1)^(3/2)]/3 + 2(x - 1)^(1/2) + k

    3.) u = sinx --> du/dx = cosx --> dx = 1/cosx du
    When x = Pi/2: u = 1
    When x = Pi/6: u = 0.5

    ∫ cos2xcosx dx
    = ∫ (Limits 1 and 0.5) cos2xcosx/cosx du
    = ∫ (Limits 1 and 0.5) cos2x du
    = ∫ (Limits 1 and 0.5) 1 - 2sin^2x du
    = ∫ (Limits 1 and 0.5) 1 - 2u^2 du
    = [u - (2u^3)/3] (Limits 1 and 0.5)
    = (1 - 2/3) - (0.5 - 1/12)
    = 12/12 - 8/12 - 6/12 + 1/12
    = -1/12
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    thanks a lot people!!
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    firstly, this is blatantly an attempt to make people do ure homework for u :O those questions are, like, the easiest in the integration by substitution chapter...

    anyway, here is my solution. i dont think i fully agree with either of the two above... so dnno...

    1. u^2 = x + 1, x = u^2 + 1
    dx/du = 2u
    ∫ (u² - 1)/u * 2u du
    ∫2u² - 2 du
    2/3u³ - 2u
    2/3(x+1)^3/2 - 2(x+1)^1/2

    2. u = x-1, x = u + 1, du/dx = 1, du = dx
    ∫u^1/2 + u^-1/2 du
    2/3u^3/2 + 2u^1/2
    2/3(x-1)^3/2 + 2(x-1)^1/2

    3. du/dx = cox x, cos 2x = 1 - 2sin²x = 1 - 2u²
    ∫1-2u²du
    [u-2/3u³]
    [sin x - 2/3sin ³ x] limits
    (sin pi/2 - 2/3(sin pi/2)³) - (sin pi/6 - 2/3sin³pi/6)
    (i dont have a calculator on me)
 
 
 
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