How to calculate the integral?

Question

#int_0^(pi/3)##sinx * e^(-3x)#dx

1799 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-05T14:07:20

#I=1/10[1-(3sqrt3+1)/(2e^pi)]#
#I~~-2.214069#

Note:

#color(red)(inte^(ax)sinbxdx=(e^(ax))/(a^2+b^2)(asinbx-bcosbx)+c#

Take,#a=-3and b=1.#

#inte^(-3x)sin(1x)dx=(e^(-3x))/((-3)^2+1^2)(-3sinx- 1cos(1x))+c#

#color(blue)(inte^(-3x)sinxdx=(e^(-3x))/(10)(-3sinx-cosx)+c...to (1)#

Here,

#I=int_0^(pi/3) sinx *e^(-3x)dx#

#I=int_0^(pi/3)(e^(-3x))sinxdx#

Using #(1)#,we get

#I=[(e^(-3x))/(10)(-3sinx-cosx)]_0^(pi/3)#

#I=[(e^(-3xxpi/3))/10(-3sin(pi/3)-cos(pi/3)]-[e^0/10(-3(0)-1)]#

#I=[e^-pi/10(-3xxsqrt3/2-1/2)]-[1/10(-1)]#

#I=e^-pi/10(-(3sqrt3+1)/2)+1/10#

#I=1/10[1-(3sqrt3+1)/(2e^pi)]#

#I~~1/10[1-23.14069]#

#I~~1/10[-22.14069]#

#I~~-2.214069#