Definite Integral of Trig Function

[Jimbo57]

Homework Statement

∫sin2xcosx x=0,pi/4

Homework Equations

The Attempt at a Solution

By double angle magic:
∫sin2xcosx=
∫2sinxcos^2x
u=cosx
du=-sinxdx
-2∫u^2
-2(u^3)/3
-2cos^3x/3
= -1/(3√2) for x=pi/4

How does that look folks?

 

[tiny-tim]

Hi Jimbo57!

(have a pi: π and try using the X² button just above the Reply box )

Fine down to …

-2cos^3x/3
= -1/(3√2) for x=pi/4

… what about for x = 0 ?

 

[Jimbo57]

Hi Jimbo57!

(have a pi: π and try using the X² button just above the Reply box )

Fine down to …


… what about for x = 0 ?

Lol woops, I'm used to ignoring the 0 with non trig integrals.

New answer:

(-1+2√2)/3√2

Thanks for the help Tiny-Tim! (and the pi)