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    int (x+1)^0.5/(x+5)dx
    limit 8 to 3
    is
    pi+2-5arctan(3/2)


    use x=sint
    int 1/(x+(1-x^2)^0.5)dx
    limit 1 to 0
    pi/4

    I think i did it
    but i have forgotten
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    I posted somthing on my website about the solution to your second integral, there is a nice method you can use to solve it: http://michaelmgs.f2o.org/maths_32.html.

    For the first, I am assuming you mean (x+1) rather than (x+a); you need to apply the substitution u^2 = x+1.
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    (Original post by totaljj)
    show
    int (x+1)^0.5/(x+5)dx
    limit 8 to 3
    is
    pi+2-5arctan(3/2)


    use x=sint
    int 1/(x+(1-x^2)^0.5)dx
    limit 1 to 0
    pi/4

    I think i did it
    but i have forgotten
    I got pi+2-4arctan(3/2) for the first one
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    do they come from past papers?
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    (Original post by keisiuho)
    do they come from past papers?
    These are from exercises in the heinmann book, I think it is exercise 3C which has about 70-80 integration questions the last few of which are quite difficult.
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    (Original post by totaljj)
    show
    int (x+1)^0.5/(x+5)dx
    limit 8 to 3
    is
    pi+2-5arctan(3/2)


    use x=sint
    int 1/(x+(1-x^2)^0.5)dx
    limit 1 to 0
    pi/4

    I think i did it
    but i have forgotten
    There is another way without using the method above for the second question. But it is quite complicated
    you will get:
    INT cos/(sin + cos)
    = INT cos*(sin - cos)/[(sin + cos)(sin - cos)]
    = INT (cost sint - cos^2 t)/ (-cos2t)
    = INT [0.5 sin2t - 0.5(1+cos2t)] / (-cos2t)
    = INT -0.5 tan2t + 0.5 sec2t + 0.5
    As you know the integrals of tan2t and sec2t, the answer can be found.
    btw, can I use the results of integrals of tan2t and sec2t directly without proof in the exam?
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    (Original post by keisiuho)
    There is another way without using the method above for the second question. But it is quite complicated
    you will get:
    INT cos/(sin + cos)
    = INT cos*(sin - cos)/[(sin + cos)(sin - cos)]
    = INT (cost sint - cos^2 t)/ (-cos2t)
    = INT [0.5 sin2t - 0.5(1+cos2t)] / (-cos2t)
    = INT -0.5 tan2t + 0.5 sec2t + 0.5
    As you know the integrals of tan2t and sec2t, the answer can be found.
    btw, can I use the results of integrals of tan2t and sec2t directly without proof in the exam?
    Yeah, that looks like a good method. I think you can use the integrals without proof, I think they are in the formula book and can be quoted from there.
 
 
 
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