STRACT: How to solve a tricky integration problem using multiple substitutions?

  • Thread starter Thread starter 2h2o
  • Start date Start date
  • Tags Tags
    Integration
2h2o
Messages
51
Reaction score
0

Homework Statement



Evaluate the indefinite integral:

Homework Equations



\int\frac{6}{x\sqrt{25x^2-1}} dx

The Attempt at a Solution



I've tried a number of different substitutions, but this last one gets me the closest, I think.

\int\frac{6}{x\sqrt{25x^2-1}} dx = \frac{6}{5}\int\frac{1}{x\sqrt{x^2-\frac{1}{25}}} dx = \frac{6}{5}\int\frac{x}{x^2\sqrt{x^2-\frac{1}{25}}} dx


let u=\sqrt{x^2-\frac{1}{25}

then

du=\frac{x dx}{\sqrt{x^2-\frac{1}{25}}}}}

This is where I want to substitute in, but it doesn't go. \frac{x}{\sqrt{x^2-\frac{1}{25}}}}}}dx fits in, except for the x^2 in the denominator which doesn't work with my substitution. I don't see where to go from here, or what I should have done differently.

Thanks!
 
Physics news on Phys.org
Try a trig substitution x=cosecθ
 
2h2o said:

Homework Statement



Evaluate the indefinite integral:

Homework Equations



\int\frac{6}{x\sqrt{25x^2-1}} dx

The Attempt at a Solution



I've tried a number of different substitutions, but this last one gets me the closest, I think.

\int\frac{6}{x\sqrt{25x^2-1}} dx = \frac{6}{5}\int\frac{1}{x\sqrt{x^2-\frac{1}{25}}} dx = \frac{6}{5}\int\frac{x}{x^2\sqrt{x^2-\frac{1}{25}}} dx


let u=\sqrt{x^2-\frac{1}{25}

then

du=\frac{x dx}{\sqrt{x^2-\frac{1}{25}}}}}

This is where I want to substitute in, but it doesn't go. \frac{x}{\sqrt{x^2-\frac{1}{25}}}}}}dx fits in, except for the x^2 in the denominator which doesn't work with my substitution. I don't see where to go from here, or what I should have done differently.

Thanks!

You're almost done: after substituting x^2=u^2+1/25 in the last integral you obtained, just bring 5 back into the integrand and then you need another substitution, basically, think of u=\sqrt(5)*cosh(x). Then you'll be left with a simple integral which is neccessarily required to be given a form like e^t/(1+e^2t) dt. From this point, you need one more substitution and then you are done by going back to x as you begin from this last one, pass t and u to definitely hit x again.

AB
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top