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Thread: 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:
    $\displaystyle sin(a)sin(b) = \frac{1}{2}(cos(a - b) - cos(a + b))$

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

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

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

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

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

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

    You can now do the full problem.
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