- #1

chwala

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## Homework Statement

using ## u= sin 4x## find the exact value of ##∫ (cos^3 4x) dx##[/B]

## Homework Equations

## The Attempt at a Solution

## u= sin 4x## [/B]on integration

**##u^2/2=-cos4x/4 ## , →##-2u^6={cos 4x}^3 ##**...am i on the right track because now i end up with ##∫{{-2u^6}/{4.-2u^2}}du## or should i use

##du=4cos 4x dx ## to end up with ## 0.25 ∫ cos^24x du## which looks wrong to me