Is this the only way of finding this Integral?

[Lorena_Santoro]

https://youtu.be/Q2V6kMSwmFU

[Guest]

The integral is [tex]\int_0^{\pi/4} cos^3(2x)dx[/tex].
You ask "Is this the only way to do this integral" but I see no solution shown!

In any case, this is how I would do it:
Write [tex]cos^3(2x)[/tex] as [tex](cos^2(2x))(cos(2x))= (1- sin^2(2x))(cos(2x))[/tex] and use the substitution [tex]u= sin(2x)[/tex] so that [tex]du= 2 cos(2x)dx[/tex] and [tex]cos(2x)dx= \frac{1}{2}du[/tex].

When x= 0, u= sin(0)= 0 and when [tex]x= \pi/4[/tex] [tex]u= sin(\pi/4)= \sqrt{2}/2[/tex].

The integral becomes [tex]\frac{1}{2}\int_0^{\sqrt{2}/2} (1- u^2)du= \left[u- \frac{u^3}{3}\right]_0^{\sqrt{2}/2}=[/tex][tex]\frac{\sqrt{2}}{2}- \frac{2\sqrt{2}}{24}=[/tex][tex]\left(\frac{6}{12}- \frac{1}{12}\right)\sqrt{2}=[/tex][tex]\frac{5\sqrt{2}}{12}[/tex].

[Guest]

I like your solution