# A tough integral

Gold Member

## Main Question or Discussion Point

There was a tough integral in my math exam that I failed to solve it and so left it blank.After that I used maple's tutor to learn how to solve it.I understood all steps but the first.My problem is that I don't know what function of x, u is.

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

thanks

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.

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'Tis a tough one

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.

Gold Member
I did as you said but I got integral below which doesn't seem to be convertable to the right side of the equality in my first post.

$${2} \int \! \frac {1-{t}^{2}} {{t}^{4}+{2}{t}^{3}+{2}{t}-{1}} {dt}$$

You may need to click on "Show Steps" in the upper right hand corner
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

Ooooops double post
Sorry

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Gold Member
I think our professor thought we are professors too. Any way.I got it.Thanks

I think our professor thought we are professors too. Reconsidered post deleted

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I think the most obvious method is, after applying the substitution x=3sin(t), just multiply numerator and denominator by sin(t)+cos(t) and then apply the double angle formulas.

I remember solving this sort of problems in highschool FP3.....best way to do is to remove square root by a suitable substitution.

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

$$\int \frac{\cos x}{\sin x - \cos x}\; dx = \frac{1}{2} \int \frac{\cos x + \sin x}{\sin x - \cos x} -\frac{\sin x - \cos x}{\sin x - \cos x} \;dx$$

which is easily

$$\frac12\left(\ln|\sin x - \cos x| - x)+C$$

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