What is the correct solution for this integral?

  • Thread starter Thread starter alba_ei
  • Start date Start date
  • Tags Tags
    Integral
alba_ei
Messages
38
Reaction score
1

Homework Statement



\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta

The Attempt at a Solution



i took u = ae^\theta-b so e^\theta = \frac{u + b}{a} then i substituded back into the integral and iget this

\int \frac{u + b + b}{u} \, du

\int du +\int \frac{2b}{u} \, du

= u \du + 2b \ln u +C

= u + 2b \ln u +C

= ae^\theta-b + 2b\ln (ae^\theta-b)

but the answer of the book is
\int \frac{ae^\theta+b}{ae^\theta-b} \, d\theta = 2\ln (ae^\theta-b) - \theta + C
what did i do wrong?
 
Last edited:
Physics news on Phys.org
You didn't subsitute properly. You have to change dtheta too.
 
write d0 as what it should equal to du
for example if u=x^2
du=2*xdx
 
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