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Math Help - Integration

  1. #1
    413
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    Integration

    how do u evaluate this?

    integral [3x (sinx/cos^4x)] dx
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    Quote Originally Posted by 413 View Post
    how do u evaluate this?

    integral [3x (sinx/cos^4x)] dx
    Use Integration by Parts,
    Let,
    u=3x and v'=sin x/cos^4 x
    Then,
    u'=3 and v=INTEGRAL sin x/cos ^4 x dx
    But how do we find that integral?
    We have, (multiply by negative)
    -INTEGRAL -sin x/cos^4 x dx
    Let, t=cos x then t'=-sin x
    Apply substitution rule,
    -INTEGRAL t'/t^4 dx=INTEGRAL 1/t^4 dt
    Thus, by using t^{-4} and exponent rule we have,
    -1/3*t^{-3}
    Substitute back,
    (-1/3)cos^{-3} x
    Thus by parts,
    -xcos^3 t+(1/3)INTEGRAL 1/cos^3 x dx
    The problem is how to deal with that integral.
    I believe that if you can express it then you should write,
    sec x*sec^x
    Apply integration by parts.
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  3. #3
    413
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    the method that you're using looks a little complicated, i got some hints on this question, it is to first find ∫ sec^3(x) dx

    ∫ sec^3(x) dx
    = ∫ sec x d tan x
    = sec x tan x - ∫ tan x d sec x
    = sec x tan x - ∫ sec x tan^2(x) dx
    then i kinda got stuck here, how do u further simplify this?
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  4. #4
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    Quote Originally Posted by 413 View Post
    the method that you're using looks a little complicated, i got some hints on this question, it is to first find ∫ sec^3(x) dx

    ∫ sec^3(x) dx
    = ∫ sec x d tan x
    = sec x tan x - ∫ tan x d sec x
    = sec x tan x - ∫ sec x tan^2(x) dx
    then i kinda got stuck here, how do u further simplify this?
    I see what you are saying,
    ∫ sec^2 x * tan x dx
    Let,
    u=tan x then, u'=sec^2 x
    Apply substitution rule,
    ∫ u du
    Thus,
    (1/2)u^2+C
    Substitute back,
    (1/2)tan^2 x+C
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