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Math Help - trigonometric integration

  1. #1
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    trigonometric integration

    evaluate the following integrals, showing all working

    a) int((sec^3*3x)(tan3x)) dx

    b)int((sin15x)(cos9x)) dx

    I was alright until the trig questions, could someone please get me started on these two.
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  2. #2
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    Quote Originally Posted by samdmansam View Post
    evaluate the following integrals, showing all working

    a) int((sec^3*3x)(tan3x)) dx Mr F says: Make the substitution u = cos x.

    b)int((sin15x)(cos9x)) dx Mr F says: 2 sin A cos B = sin (A + B) + sin (A - B) ......

    I was alright until the trig questions, could someone please get me started on these two.
    ..
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  3. #3
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    He4llo, samdmansam!

    Another approach to (a) . . .


    a)\;\;\int \sec^3\!3x \tan3x\,dx

    \text{Let: }\:u \:=\:\sec3x\quad\Rightarrow\quad du \:=\:3\sec3x\tan3x\,dx

    \text{We have: }\;\int\underbrace{\sec^2\!3x}_{u^2}\underbrace{(\  sec3x\tan3x\,dx)}_{\frac{1}{3}du} \;=\;\frac{1}{3}\int u^2\,du


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    When we have: . \int\sec^m\!x\tan^n\!x\,dx
    . . there is a procedure we can follow.

    There are three cases to consider . . .


    [1] Even secant, any tangent.

    Example: . \int\sec^4\!x\tan^3\!x\,dx

    \text{"Peel off" }\sec^2\!x\text{ for our }du.

    We have: . \int\sec^2\!x\,\tan^3\!x\,(\sec^2\!x\,dx)
    . . . . . . . . . . \uparrow
    . . . . .
    change to tangents
    . . . . . . . . . \downarrow
    . . = \;\int(\tan^2\!x + 1)\tan^3\!x\,(\sec^2\!x\,dx) \;=\;\int(\tan^5\!x + \tan^3\!x)\,(\sec^2\!x\,dx) <br />


    \text{Let: }\:u = \tan x\quad\Rightarrow\quad du = \sec^2\!x\,dx

    . . And we have: . \int(u^5 + u^3)\,du



    [2] Odd tangent, any secant

    Example: . \int\sec^5\!x\tan^3\!x\,dx

    \text{"Peel off" }\sec x\tan x\text{ for our }du.

    We have: . \int\sec^4\!x\tan^2\!x\,(\sec x\tan x\,dx)
    . . . . . . . . . . . . . . \uparrow
    . . . . . . . . .
    change to secants
    . . . . . . . . . . . . \downarrow
    . . = \;\int\sec^4\!x(\sec^2\!x-1)\,(\sec x\tan x\,dx) \;=\;\int(\sec^6\!x - \sec^4\!x)\,(\sec x\tan x\,dx)


    \text{Let: }\,u \,=\,\sec x\quad\Rightarrow\quad du \,=\,\sec x\tan x\,dx

    . . And we have . \int(u^6 - u^4)\,du



    [3] Odd secant, even tangent

    This is the dreaded case, where a formula must be known
    . . or it must be derived.

    . . \int\sec^3\!x\,dx \;=\;\frac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] + C

    and a reduction formula is also recommended.

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