Simpler Solution for Cumbersome Trig Integral?

[Nebuchadnezza]

So this is not an homework question, and I have solved the integral...

Just it took a lot of time and was very tiresome.

The integral

\int_{\pi/2}^{\pi/6} \sin(2x)^3\cdot \cos(3x)^2 dx
Can be expressed on the form \Large \left( \frac{a}{b} \right)^b where a and b are integers.

Evaluate \sqrt{a^b+b^a-1}

A quick outline on how I did it. I mainly used two formulas

\sin(A)cos(B)=\sin(A+B)+\sin(A-B)

\sin(A)\sin(B)\sin(C)=\frac{1}{4} \left[ \sin(A+B-C) - \sin(A-B-C) + \sin(A-B+C) - \sin(A+B+C) \right]

A short outline, the steps omitted are those that take a lot of time...I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx}
I = \int\limits_{\pi /6}^{\pi /2} {{{\left[ {\sin \left( {2x} \right)\cos \left( {3x} \right)} \right]}^2}\sin \left( {2x} \right)dx}
I = \frac{1}{4}\int\limits_{\pi /6}^{\pi /2} {{{\left[ {\sin \left( {2x - 3x} \right) + \sin \left( {2x + 3x} \right)} \right]}^2}\sin \left( {2x} \right)} dx
I = \frac{1}{4}\int\limits_{\pi /6}^{\pi /2} {\left[ {\sin {{\left( x \right)}^2} - 2\sin \left( x \right)\sin \left( {5x} \right) + \sin {{\left( {5x} \right)}^2}} \right]\sin \left( {2x} \right)} dx
I = \frac{1}{{16}}\int\limits_{\pi /6}^{\pi /2} {6\sin \left( {2x} \right) - 3\sin \left( {4x} \right) - 2\sin \left( {6x} \right) + 3\sin \left( {8x} \right) - \sin \left( {12x} \right)dx}
I = \frac{1}{{16}}\left[ { - \frac{6}{2}\cos \left( {2x} \right) + \frac{3}{4}\cos \left( {4x} \right) + \frac{2}{6}\sin \left( {6x} \right) - \frac{3}{8}\cos \left( {8x} \right) + \cos \left( {12x} \right)} \right] + C
I = \frac{1}{{16}}\left[ {\left( { - 3\cos \left( \pi \right) + \frac{3}{4}\cos \left( {2\pi } \right) + \frac{1}{3}\sin \left( {3\pi } \right) - \frac{3}{8}\cos \left( {4\pi } \right) + \cos \left( {6\pi } \right)} \right) - \left( { - 3\cos \left( {\frac{\pi }{3}} \right) + \frac{3}{4}\cos \left( {\frac{{2\pi }}{3}} \right) + \frac{1}{3}\sin \left( \pi \right) - \frac{3}{8}\cos \left( {\frac{{4\pi }}{3}} \right) + \cos \left( {2\pi } \right)} \right)} \right]
I = \frac{1}{{16}}\left[ {\left( {3 + \frac{3}{4} + 0 - \frac{3}{8} + 1} \right) - \left( { - \frac{3}{2} - \frac{3}{8} + 0 + \frac{3}{8}\left( { - \frac{1}{2}} \right) + 1} \right)} \right]
I = \frac{1}{{16}}\left[ {\left( {\frac{{35}}{8}} \right) - \left( { - \frac{{11}}{{16}}} \right)} \right] = \frac{1}{{16}}\left[ {\frac{{70 + 11}}{{16}}} \right] = \frac{{81}}{{256}} = {\left( {\frac{3}{4}} \right)^4}
I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} = {\left( {\frac{3}{4}} \right)^4}
\sqrt {{3^4} + {4^3} - 1} = \sqrt {81 + 64 - 1} = \sqrt {144} = \fbox{12}

Step 4 to 5 took me forever to figure out. Had to "invent" the formula above, after many failed attempts at simplifying the expression.

So anyone have a simpler, easier solution?

 

[micromass]

An alternate method might be to apply the formulas

\cos^2(3x)=\frac{1+\cos(6x)}{2}

and

\sin^3(2x)=\frac{3\sin(2x)-\sin(6x)}{4}

This has the benifit that you don't need to work with exponents. But it's probably as much calculation...

 

[I like Serena]

Funny! I did exactly the same thing!

For reference you might use:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities
where you can find these formulas (sort of).
Using this list of trig formulas eliminates a lot of figuring out.

But a very quick way to do it, is this one, although you might consider it cheating:
http://www.wolframalpha.com/input/?i=\int_{\pi%2F6}^{\pi%2F2}+\sin%282x%29^3\cdot+\cos%283x%29^2+dx

 

[andylu224]

You've got a few errors in your integration. Integrating -2sin6x should be 1/3 cos6x and -sin12x should be 1/12 cos12x. The final answer seems alright surprisingly.