MHB What's the best strategy to solving this Integral in 3 minutes?

[Lorena_Santoro]

 

[topsquark]

Please do not double post here, either.

-Dan

 

[HOI]

Separate cos^3(2x) as (cos^2(2x))cos(2x)= (1- sin^2(2x))cos(2x). Now use the substitution u= sin(2x) so that du= 2cos(2x)dx, cos(2x)dx= (1/2)du. When x= 0, u= sin(0)= 0 and when x= 4, u= sin(8). The integral becomes $$\frac{1}{2}\int_0^{sin(8)}(1- u^2)du=\left[u- \frac{u^3}{3}\right]_0^{sin(8)}$$$$= sin(8)- \frac{sin^3(8)}{3}$$.

 

[skeeter]

$x = \dfrac{\pi}{4} \implies \text{ upper limit }, u= 1$

 

[Lorena_Santoro]

Yes, and thank you!