- #1

ShayanJ

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[tex] \int \! \frac{dx}{ x-\sqrt {9-{x}^{2}}}=\int \!4\,{\frac {u}{-1+{u}^{4}+2\,{u}^{3}+2\,u}} {du}[/tex]

thanks

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- Thread starter ShayanJ
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- #1

ShayanJ

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[tex] \int \! \frac{dx}{ x-\sqrt {9-{x}^{2}}}=\int \!4\,{\frac {u}{-1+{u}^{4}+2\,{u}^{3}+2\,u}} {du}[/tex]

thanks

- #2

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First try x=3sin(t). That should eliminate the square root and that will give you a integral of a function containing sin(t) and cos(t).

Then, try the substitution u=tan(t/2). That should give you the right-hand side.

Then, try the substitution u=tan(t/2). That should give you the right-hand side.

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- #3

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The 1st substitution x=3sin t will simplify to

Integral cos t / [ sin t - cos t ] dt

This is perfect for the Tangent Half Angle Method

sometimes called Weierstrass Method.

Let u = tan (t/2)

One needs some experience to arrive at the 4th order polynomial

you have but it is the correct one. Then partial fraction expansion

is called for.

See the near full solution at www.wolframalpha.com

http://www.wolframalpha.com/input/?i=integral++cos+x+/+[+%28sin+x+-+cos+x%29+]++dx

[ You may need to cut and paste this link ]

One still needs to back substitute.

- #4

ShayanJ

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[tex] {2} \int \! \frac {1-{t}^{2}} {{t}^{4}+{2}{t}^{3}+{2}{t}-{1}} {dt} [/tex]

- #5

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to see the step by step solution.

Tangent Half Angle Method converts a rational trig integrand to a rational algebraic integrand.

But you need some experience as I mentioned as the details do look complex.

Study the Wolfram solution and you should find your error

Note it is still a long way to the final Antiderivative.

Partial Fraction Expansion is required and then back substitution

Keep at it. It will feel good to do such an involved integration

- #6

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Ooooops double post

Sorry

Sorry

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- #7

ShayanJ

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I think our professor thought we are professors too.

Any way.I got it.Thanks

Any way.I got it.Thanks

- #8

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I think our professor thought we are professors too.

Reconsidered post deleted

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- #9

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- #11

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I don't think anyone saw the shortcut: once you do the trig substiution, do this

[tex]\int \frac{\cos x}{\sin x - \cos x}\; dx =

\frac{1}{2} \int \frac{\cos x + \sin x}{\sin x - \cos x}

-\frac{\sin x - \cos x}{\sin x - \cos x} \;dx[/tex]

which is easily

[tex]\frac12\left(\ln|\sin x - \cos x| - x)+C[/tex]

[tex]\int \frac{\cos x}{\sin x - \cos x}\; dx =

\frac{1}{2} \int \frac{\cos x + \sin x}{\sin x - \cos x}

-\frac{\sin x - \cos x}{\sin x - \cos x} \;dx[/tex]

which is easily

[tex]\frac12\left(\ln|\sin x - \cos x| - x)+C[/tex]

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