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Math Help - Please walk me through this integral

  1. #1
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    Please walk me through this integral

    (sin (x))(tan^2 (x))

    Please help me understand the process of evaluating this integral. Thanks,

    Kim
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  2. #2
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    Quote Originally Posted by Kim Nu View Post
    (sin (x))(tan^2 (x))

    Please help me understand the process of evaluating this integral. Thanks,

    Kim

    There may be a faster way but this is how I would do it.
    rewirte the integrand as

    \sin(x)\cdot \frac{\sin^{2}(x)}{\cos^2(x)}=\sin^{3}(x)\sec^{2}(  x)

    Now we will integrate by parts with

    u=\sin^{3}(x) \to du = 3\sin^{2}(x)\cos(x)

    dv=\sec^{2}(x) \to v=\tan(x)

     <br />
\sin^{3}(x)\tan(x)-\int 3\sin^{2}(x)\cos(x)\tan(x)dx<br />

    rewrite the integrand as

     <br />
\sin^{3}(x)\tan(x)-\int 3\sin^{3}(x)dx = \sin^{3}(x)\tan(x)-\int 3\sin(x)(1-\cos^2(x))dx=
    \sin^{3}(x)\tan(x)-3\int \sin(x)dx+3\int \sin(x)\cos^2(x))dx <br />

    The last integral can be done with a u sub u=cos(x)

    and we finally get

    \sin^{3}(x)\tan(x)+3\cos(x)-\cos^3(x) <br />

    Maybe there is a faster way....

    Anyone...

    I hope this helps
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  3. #3
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    Quote Originally Posted by Kim Nu View Post
    (sin (x))(tan^2 (x))

    Please help me understand the process of evaluating this integral. Thanks,

    Kim

    Okay this works even faster

    \int \sin(x)\tan^{2}(x)

    by parts again

    u=\tan^{2}(x) \to du=2\tan(x)\sec^{2}(x)dx

    dv=\sin(x) \to v= -\cos(x)

    Now we get

    -\cos(x)\tan^{2}(x)-\int(-\cos(x))(2\tan(x)\sec^{2}(x))dx

    Cleaning up a bit gives

    -\cos(x)\tan^{2}(x)+2\int\frac{\sin(x)}{\cos^{2}(x)  }dx

    Now just let u=cos(x) to finish

    So we end up with

    -\cos(x)\tan^{2}(x)+\frac{2}{\cos(x)}=

    -\cos(x)\tan^{2}(x)+2\sec(x)+C
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  4. #4
    Super Member wingless's Avatar
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    This approach uses only one substitution:

    \int\sin x \tan^2 x~dx

    \int \frac{\sin^3 x}{\cos^2 x}~dx

    \int \frac{\sin x (1-\cos^2 x)}{\cos^2 x}~dx

    \int \frac{\sin x}{\cos^2 x} - \sin x~dx

    \int \frac{\sin x}{\cos^2 x}~dx + \cos x + C

    Let u = \cos x, then du = -\sin x~dx

    -\int \frac{du}{u^2} + \cos x + C

    \frac{1}{u} + \cos x + C

    \frac{1}{\cos x} +\cos x + C
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  5. #5
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    Hey thanks everyone.

    How about this one: (tan^2(x)) / cos(x)

    Thanks, Kim
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by Kim Nu View Post
    Hey thanks everyone.

    How about this one: (tan^2(x)) / cos(x)

    Thanks, Kim
    This integral doesn't have an easy solution, it involves the use of hyperbolic sine or cosine functions. Do you know them?
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  7. #7
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Kim Nu View Post
    Hey thanks everyone.

    How about this one: (tan^2(x)) / cos(x)

    Thanks, Kim
    Let u=\sin(x) then: <br />
\int {\frac{{\sin ^2 \left( x \right)}}<br />
{{\cos ^3 \left( x \right)}}dx}  = \int {\frac{{u^2 }}<br />
{{\left( {1 - u^2 } \right)^2 }}du}  <br />

    note that: <br />
\int {\frac{{u^2 }}<br />
{{\left( {1 - u^2 } \right)^2 }}du}  =  \tfrac{1}<br />
{2}\int {u \cdot \left( {\frac{1}<br />
{{1 - u^2 }}} \right)^\prime  du} <br />
(1)

    Apply parts in (1)

    And remember that: \frac{1}{(1-u^2)}=\frac{1}{2}\cdot{\left(\frac{1}{1-u}+\frac{1}{1+u}\right)}
    Last edited by PaulRS; May 29th 2008 at 12:04 PM.
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  8. #8
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    A slight improvement on wingless's solution . . .

    This approach uses only one substitution:

    \int\sin x \tan^2\!x\,dx \;=\;\int \frac{\sin^3\!x}{\cos^2\!x}\,dx \;=\;\int \frac{\sin x (1-\cos^2\!x)}{\cos^2\!x}\,dx \;= . \int\left(\frac{\sin x}{\cos^2\!x} - \sin x\right)\,dx

    We have: . \int\left(\frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x} - \sin x\right)\,dx

    . . . . . . \;\;=\;\;\int(\sec x\tan x - \sin x)\,dx \;=\;\boxed{\sec x + \cos x + C}

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  9. #9
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    Hello, Kim!

    Are you familiar with this formula?

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


    How about this one: . \int\frac{\tan^2\!x}{\cos x}\,dx

    We have: . \int\sec x\tan^2\!x\,dx \;=\;\int\sec x(\sec^2\!x - 1)\,dx \;=\;\int\left(\sec^3\!x - \sec x\right)dx

    Then: . \frac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] - \ln|\sec x + \tan x| + C

    . . . = \;\frac{1}{2}\sec x\tan x - \frac{1}{2}\ln|\sec x + \tan x| + C

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