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Math Help - integration problem

  1. #1
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    integration problem

    I have been doing couple of integration by parts and when I got to trig. integrals I get confuse, specially this problem.

    Integral cos x^2 tan x^3 dx

    I'm used to problems that has tan with sec^2's, cos with cos^2's etc
    but not different trig functions such as cos and tan.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ff4930 View Post
    I have been doing couple of integration by parts and when I got to trig. integrals I get confuse, specially this problem.

    Integral cos x^2 tan x^3 dx

    I'm used to problems that has tan with sec^2's, cos with cos^2's etc
    but not different trig functions such as cos and tan.
    i'm pretty sure that \cos \left( x^2 \right) \tan \left( x^3 \right) is not integrable in terms of elementary functions

    did you mean \int \cos^2 x \tan^3 x~dx ?? those are two very different things. be specific as to what you want, and careful
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  3. #3
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    Yes, I apologize it is cos^2x and tan^3x.
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  4. #4
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    \cos^2 x \tan^3 x = \cos ^2 x \cdot \frac{\sin^3 x}{\cos ^3 x} = \frac{(1-\cos^2 x)\sin x}{\cos x}

    Let t=\cos x and make subsitution.
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  5. #5
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    Quote Originally Posted by ff4930 View Post
    Yes, I apologize it is cos^2x and tan^3x.
    Let me practice on integration.

    INT.[cos^2(X) tan^3(X)]dX
    = INT.[cos^2(X) sin^3(X)/cos^3(x)]dX
    = INT.[ sin^3(X) /cosX]dX
    = INT.[{1 -cos^2(X)}sinX/cosX]dX
    = INT.[sinX/cosX -sinXcosX]dX
    = INT.[tanX -(1/2)sin(2X)]dX
    = INT.[tanX]dX -(1/2)INT.[sin(2X)]dX
    = ln[secX] -(1/2)(1/2)INT.[sin(2X)](2dX)
    = ln[secX] -(1/4)[-cos(2X)]
    = ln[secX] +(1/4)cos(2X) +C --------------answer.
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