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Math Help - [SOLVED] definite intergral problem

  1. #1
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    Question [SOLVED] definite intergral problem

    Hi - I'd appreciate it if someone could point out where i'm going wrong on this one:

    Intergrate within the limits of pi/2 and 0

    (sinx)^3 * (cosx)^2 dx

    Ok, use the identity (sinx)^2 = 1 - (cosx)^2

    and I get sinx(cosx)^2 - sinx(cosx)^4

    Intergrate and i get:

    -1/3(cosx)^3 + 1/5(cosx)^5

    Evaluate within the limits and I get zero as cos pi/2 equals zero.

    This isnt the answer i get on an itegration calculator???

    thanks D
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  2. #2
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    Quote Originally Posted by dojo View Post
    Hi - I'd appreciate it if someone could point out where i'm going wrong on this one:

    Intergrate within the limits of pi/2 and 0

    (sinx)^3 * (cosx)^2 dx

    Ok, use the identity (sinx)^2 = 1 - (cosx)^2

    and I get sinx(cosx)^2 - sinx(cosx)^4

    Intergrate and i get:

    -1/3(cosx)^3 + 1/5(cosx)^5

    Evaluate within the limits and I get zero as cos pi/2 equals zero.

    This isnt the answer i get on an itegration calculator???

    thanks D


    But \cos 0=1 !! So the answer must be:

    \frac{1}{3}[\cos^3(\pi\slash 2)-\cos^3(0)]+\frac{1}{5}[\cos^5(\pi\slash 2)-\cos^5(0)]=-\frac{1}{3}(-1)+\frac{1}{5}(-1)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}

    Tonio
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  3. #3
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    Thanks - Schoolboy error . I've had my nose in my books too long today!
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by dojo View Post
    Hi - I'd appreciate it if someone could point out where i'm going wrong on this one:

    Intergrate within the limits of pi/2 and 0

    (sinx)^3 * (cosx)^2 dx

    Ok, use the identity (sinx)^2 = 1 - (cosx)^2

    and I get sinx(cosx)^2 - sinx(cosx)^4

    Intergrate and i get:

    -1/3(cosx)^3 + 1/5(cosx)^5

    Evaluate within the limits and I get zero as cos pi/2 equals zero.

    This isnt the answer i get on an itegration calculator???

    thanks D
    I will show you the steps involved. I hope you can plug in the limits and calculate:

    \int sin^{3}x cos^{2}x dx

    = \int sinx \times cos^{2}x \times sin^{2}x  dx

    = \int sinx \times cos^{2}x (1-cos^{2}x)  dx

    Now use substitution rule:

    let  u = cosx , therefore du = -sinx dx that gives  dx = \frac {-du}{sinx}

    your integral becomes:

    =  -\int u^2 (1-{u^2})  du

    = - \int ({u^2}-{u^4})  du

    = \int {u^4}du - \int {u^2}du

    = \frac{u^5}{5} - \frac{u^3}{3}

    substitute u = cosx to get

    =  [\frac {cos^{5}x}{5}] - [\frac{cos^{3}x}{3}]


    then evaluate these fractions with the limits you have...
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