How can the integral of a complex function be simplified?

  • Thread starter Thread starter gaobo9109
  • Start date Start date
  • Tags Tags
    Integration
gaobo9109
Messages
68
Reaction score
0

Homework Statement



\int sin e^{-x}+e^x cos e^{-x}\,dx

Find the integral above

Homework Equations





The Attempt at a Solution



I tried substituting u=e^{-x}, but i get \int \frac{sin u}{u}+\frac{cos u}{u^2} \,du, which is non-integrable function.
 
Last edited:
Physics news on Phys.org
What is the derivative of ##e^x\cos(e^{-x})##?

ehild
 
Thanks. I didn't notice that
 
You are welcome.

ehild
 
ehild said:
What is the derivative of ##e^x\cos(e^{-x})##?

ehild

A second way is to try to transform the cos into a sin function. That leads to the hope that an integration by part will work by starting with

e^x cos (e^{-x}) = \frac{d e^x}{dx} \cos (e^{-x})

and sure enough that works.
 
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