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Math Help - integration by substitution

  1. #1
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    integration by substitution

    If someone could please help me solve some of these, because I'm getting stuck on all of them:

    1. ∫ sin5xsin3x dx

    2. ∫ (tanx + cot x) squared

    3. ∫ (tan x) to the sixth power

    4. ∫ (tan x)cubed times (sec x) cubed

    thanks for any help!
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  2. #2
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    Quote Originally Posted by turtle View Post
    If someone could please help me solve some of these, because I'm getting stuck on all of them:

    1. ∫ sin5xsin3x dx

    2. ∫ (tanx + cot x) squared

    3. ∫ (tan x) to the sixth power

    4. ∫ (tan x)cubed times (sec x) cubed

    thanks for any help!
    My eyes are starting to get bleary, (it's been a long day), so I will just say the first one will go much easier if you note:
    sin(a)sin(b) = \frac{1}{2}(cos(a - b) - cos(a + b))

    -Dan
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  3. #3
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    Quote Originally Posted by turtle View Post

    3. ∫ (tan x) to the sixth power
    \int \tan^6 x dx
    Write,
    \int \tan^4 x \tan^2 x dx
    \int \tan^4 x (\sec ^2 x-1) dx
    \int \tan^4 x \sec ^2 x dx - \int \tan^4 x dx
    But,
    \int \tan^4 x dx= \int \tan^2 x(\sec^2 x -1) dx=\int \tan^2 x \sec^2 xdx - \int \tan^2 x dx
    But,
    \int \tan^2 xdx=\int \sec^2 x -1 dx
    Thus,
    \int \tan^4 x \sec^2 xdx-\int \tan^2 x\sec^2 dx+\int \sec^2 x dx-\int x dx
    You can do the first do using u=\tan x.
    The third is simply \tan x.
    The last is easy.
    Thus,
    \frac{1}{5}\tan^5 x-\frac{1}{3}\tan^3 x-\tan x-\frac{1}{2}x^2+C
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  4. #4
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    Quote Originally Posted by turtle View Post

    2. ∫ (tanx + cot x) squared
    Open paranthesis,
    \int \tan^2 x+2\tan x\cot x+\cot^2 xdx
    Thus, (note that \tan x\cot x=1),

    \int \tan^2 x+1dx+\int cot^2 x +1dx

    \int \sec^2 xdx +\int \csc^2 x dx

    \tan x - \cot x +C
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  5. #5
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    Quote Originally Posted by turtle View Post

    4. ∫ (tan x)cubed times (sec x) cubed
    \tan^3 x\sec^3 x=\tan^3 x\sec^2 x \sec x
    Call,
    u'=\tan^3 x\sec^2 x
    v=\sec x
    Thus,
    u=\frac{1}{4}\tan^4 x
    v'=-\sin x\cot^2 x
    The integral,
    \int \tan^3 x\sec^2 x \sec x dx
    Can be done by parts as above,
    \frac{1}{4}\sec x\tan^4 x +\frac{1}{4} \int \tan^4 x \sin x \cot ^2 x dx
    But,
    \tan^2 x\cot^2 x=1
    Thus,
    \frac{1}{4}\sec x\tan^4 x+\frac{1}{4} \int \tan^2 x\sin x dx
    The problem reduces to finding,

    \int \tan^2 x\sin x dx = \int \frac{\sin^3 x}{\cos^2 x} dx
    Write,
    \frac{\sin^3 x}{\cos^2 x}=\frac{(1-\cos^2 x)}{\cos ^2 x} \cdot \sin x
    Use,
    t=\cos x
    t'=-\sin x
    Thus,
    \int -\frac{1-t^2}{t^2} t' dx
     \int 1- t^{-2} dt
    t+t^{-1}
    \cos x+\sec x+C

    You can now do the full problem.
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