- #1
Nebuchadnezza
- 79
- 2
So this is not an homework question, and I have solved the integral...
Just it took a lot of time and was very tiresome.
A quick outline on how I did it. I mainly used two formulas
[tex]\sin(A)cos(B)=\sin(A+B)+\sin(A-B)[/tex]
[tex]\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][/tex]
A short outline, the steps omitted are those that take a lot of time...[tex] I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} [/tex]
[tex] I = \int\limits_{\pi /6}^{\pi /2} {{{\left[ {\sin \left( {2x} \right)\cos \left( {3x} \right)} \right]}^2}\sin \left( {2x} \right)dx} [/tex]
[tex] 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 [/tex]
[tex] 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 [/tex]
[tex] 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} [/tex]
[tex] 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[/tex]
[tex] 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] [/tex]
[tex] 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] [/tex]
[tex] 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} [/tex]
[tex] I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} = {\left( {\frac{3}{4}} \right)^4} [/tex]
[tex] \sqrt {{3^4} + {4^3} - 1} = \sqrt {81 + 64 - 1} = \sqrt {144} = \fbox{12} [/tex]
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?
Just it took a lot of time and was very tiresome.
The integral
[tex] \int_{\pi/2}^{\pi/6} \sin(2x)^3\cdot \cos(3x)^2 dx[/tex]
Can be expressed on the form [itex]\Large \left( \frac{a}{b} \right)^b[/itex] where [itex]a[/itex] and [itex]b[/itex] are integers.
Evaluate [itex]\sqrt{a^b+b^a-1}[/itex]
A quick outline on how I did it. I mainly used two formulas
[tex]\sin(A)cos(B)=\sin(A+B)+\sin(A-B)[/tex]
[tex]\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][/tex]
A short outline, the steps omitted are those that take a lot of time...[tex] I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} [/tex]
[tex] I = \int\limits_{\pi /6}^{\pi /2} {{{\left[ {\sin \left( {2x} \right)\cos \left( {3x} \right)} \right]}^2}\sin \left( {2x} \right)dx} [/tex]
[tex] 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 [/tex]
[tex] 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 [/tex]
[tex] 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} [/tex]
[tex] 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[/tex]
[tex] 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] [/tex]
[tex] 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] [/tex]
[tex] 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} [/tex]
[tex] I = \int\limits_{\pi /6}^{\pi /2} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2}dx} = {\left( {\frac{3}{4}} \right)^4} [/tex]
[tex] \sqrt {{3^4} + {4^3} - 1} = \sqrt {81 + 64 - 1} = \sqrt {144} = \fbox{12} [/tex]
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?