micromass said:For the first one: factor ou the 6 and find a suitable substitution to bring the integral back to
[tex]\int{e^ydy}[/tex]
JHamm said:For the second one you can separate it into two integrals, sin(pi x) dx and -3/x dx
adimi24 said:But am still confused with this one. So far i have :
-1/4 ∫ cos(pi x) - 3Ln| x|
...
Also, this may be a silly question but am I supposed to use pi as in 3.14 or 180
Bohrok said:How did you get -1/4 ∫cos(pi x) ?
x is probably in radians, so you'll use 3.14 instead of 180 for degrees.
The two integral problems that involve e and sin are: ∫e^x*sin(x)dx and ∫e^x*sin^2(x)dx.
The general form of the solution to these integral problems is: ∫e^x*sin(x)dx = -1/2e^x(cos(x) - sin(x)) + C and ∫e^x*sin^2(x)dx = -1/4e^x(cos(2x) - 2sin(2x)) + C.
These integral problems can be solved using integration by parts. The first integral can be solved by setting u = sin(x) and dv = e^x, while the second integral can be solved by setting u = sin^2(x) and dv = e^x.
Yes, for the first integral, the pattern to follow is ∫e^x*sin(x)dx = -e^x*cos(x) + ∫e^x*cos(x)dx, and for the second integral, the technique to use is trigonometric identities to simplify the integrand.
These integral problems have various applications in physics, engineering, and other fields. For example, they can be used to model electrical circuits and analyze alternating currents, as well as to solve problems involving oscillatory motion and resonance.