- #1

- 96

- 0

## 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?

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- Thread starter Jimbo57
- Start date

- #1

- 96

- 0

∫sin2xcosx x=0,pi/4

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?

- #2

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

(have a pi: π and try using the X

Fine down to …

-2cos^3x/3

= -1/(3√2) for x=pi/4

… what about for x = 0 ?

- #3

- 96

- 0

Hi Jimbo57!

(have a pi: π and try using the X^{2}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)

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