Question

Fun integral

Answer 1

\[\int \sin (101 x)\left(\sin x\right)^{99} dx\]

Answer 2

At last someone posted a good question

Answer 3

It's too obvious once you see it.

Answer 4

the title is self contradicting

Answer 5

maybe this identity can be useful:
\[\sin \left( {101x} \right){\left( {\sin x} \right)^{99}} = \left( {\frac{{{e^{i101x}} - {e^{ - i101x}}}}{{2i}}} \right) \cdot {\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^{99}}\]

Answer 6

I haven't worked with complex exponentials before, but they are the least of our worries in this problem.

Answer 7

If I use the formula of Newton, I get this:
\[\begin{gathered}
\sin \left( {101x} \right){\left( {\sin x} \right)^{99}} = \left( {\frac{{{e^{i101x}} - {e^{ - i101x}}}}{{2i}}} \right) \cdot {\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^{99}} = \hfill \\
\hfill \\
= \sum\limits_{k = 0}^{99} {\left( {\begin{array}{*{20}{c}}
{99} \\
k
\end{array}} \right)} {\left( { - 1} \right)^{99 - k}}\left\{ {{e^{2i\left( {k + 1} \right)x}} - {e^{2i\left( {k - 100} \right)x}}} \right\} \hfill \\
\end{gathered} \]

Answer 8

Oh wow, you definitely don't need to do all that.

Answer 9

oops...
\[\begin{gathered}
\sin \left( {101x} \right){\left( {\sin x} \right)^{99}} = \left( {\frac{{{e^{i101x}} - {e^{ - i101x}}}}{{2i}}} \right) \cdot {\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^{99}} = \hfill \\
\hfill \\
= \frac{1}{{{2^{100}}}}\sum\limits_{k = 0}^{99} {\left( {\begin{array}{*{20}{c}}
{99} \\
k
\end{array}} \right)} {\left( { - 1} \right)^{99 - k}}\left\{ {{e^{2i\left( {k + 1} \right)x}} - {e^{2i\left( {k - 100} \right)x}}} \right\} \hfill \\
\end{gathered} \]

Answer 10

I will reveal the solution now...

Answer 11

@L
Wait, give people more time to see more thoughts.
Give it at least 24 hours!

Answer 12

Aw, OK then.

Answer 13

@L thanks!

Answer 14

*

Answer 15

As promised... it's just a simple trick.\[\sin(101x) (\sin x)^{99}\]\[= \sin(100x + x)(\sin x)^{99}\]\[= \sin(100x ) \cos(x)\sin^{99}(x) + \cos(100x) \sin^{100}(x)\]But observe that the derivative of \(\sin 100x\) is \(100 \cos 100x\) and the derivative of \(\sin^{100}x\) is \(100 \sin^{99}x \cos x\) in return. This hints us at the product rule.\[= \frac{1}{100 } \left(\sin^{100}x \cdot( \sin 100x) ' +(\sin^{100 }x)' \cdot \sin(100x)\right)\] \[= \frac{1}{100}\frac{d}{dx} \sin^{100}x\sin(100x)\]And if we integrate this, we should get the thing itself.\[\Rightarrow I = \frac{1}{100}\sin^{100}x \sin(100x)\]

Answer 16

Nice!