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Thread: need some explanation on integration

  1. #1
    Member Jones's Avatar
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    need some explanation on integration

    Hello,

    I'm trying to integrate the following:

    $\displaystyle \int sin^4 x\cdot cos^5 x$

    rewrite:
    $\displaystyle sin^4x(cosx\cdot cos^4 x)$

    $\displaystyle cos^2 x = (1-sin^2x)$

    $\displaystyle cos^4 x = (1-sin^2x)^2 = 1-2sin^2x+sin^4x$

    multiply together:

    $\displaystyle sin^4x-2sin^6x+sin^8x \cdot cosx$

    let u = sin x, du = cos x

    we have $\displaystyle u^4-2u^6+u^8 du$

    My question is, the derivative of u is cos x

    But we have 3 u terms. Aren't you getting a ŽduŽ for every u term?
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  2. #2
    Ted
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    Quote Originally Posted by Jones View Post
    Hello,

    I'm trying to integrate the following:

    $\displaystyle \int sin^4 x\cdot cos^5 x$

    rewrite:
    $\displaystyle sin^4x(cosx\cdot cos^4 x)$

    $\displaystyle cos^2 x = (1-sin^2x)$

    $\displaystyle cos^4 x = (1-sin^2x)^2 = 1-2sin^2x+sin^4x$

    multiply together:

    $\displaystyle sin^4x-2sin^6x+sin^8x \cdot cosx$

    let u = sin x, du = cos x

    we have $\displaystyle u^4-2u^6+u^8 du$

    My question is, the derivative of u is cos x

    But we have 3 u terms. Aren't you getting a ŽduŽ for every u term?
    Nope. only one $\displaystyle du$, here:

    Quote Originally Posted by Jones View Post
    $\displaystyle sin^4x-2sin^6x+sin^8x \cdot cosx$
    3
    You should have brackets here.

    It should be $\displaystyle {\color{red} ( } sin^4(x)-2sin^6(x)+sin^8(x) {\color{red} ) } cos(x)$

    take $\displaystyle u=sin(x)$ to get:

    $\displaystyle \int (u^4-2u^6+u^8)du$ which is easy.
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