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Calculus Level 5

0 π / 4 cos 10 x sin 20 x d x \large \displaystyle \int_{0}^{{\pi} / {4}}\cos^{10}x\sin{20x} \, dx

If the integral above is of the form A B \dfrac A B , where A A and B B are coprime positive integers, find A + B A+B .

Note : Mathematical as well as Programming solution is acceptable.


Inspiration


The answer is 202303.

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1 solution

Reynan Henry
Jan 10, 2017

define I a , b = cos a x sin b x d x I_{a,b}= \int \cos^a{x}\sin{bx}dx

I a , b = cos a x sin b x d x = c o s a 1 x ( cos x sin b x ) d x = cos a 1 x ( sin ( b + 1 ) x + sin ( b 1 ) x ) d x = cos a 1 x sin ( b + 1 ) x d x + cos a 1 x sin ( b 1 ) x d x = I a 1 , b + 1 + I a 1 , b 1 \begin{aligned}I_{a,b}&= \int \cos^a{x}\sin{bx}dx \\&= \int cos^{a-1}x(\cos{x}\sin{bx})dx \\&= \int \cos^{a-1}x(\sin{(b+1)x}+\sin{(b-1)x})dx\\&=\int \cos^{a-1}x\sin{(b+1)x}dx+\int \cos^{a-1}x\sin{(b-1)x}dx \\&= I_{a-1,b+1}+I_{a-1,b-1}\end{aligned}

I 0 , b = sin b x d x = 1 b cos b x I_{0,b}= \int \sin{bx}dx = -\frac{1}{b}\cos{bx}

the problem is I 10 , 20 I_{10,20}

I 10 , 20 = I 9 , 19 + I 9 , 21 = I 8 , 18 + 2 I 8 , 20 + I 8 , 22 = I 7 , 17 + 3 I 7 , 19 + 3 I 7 , 21 + I 7 , 23 = I 6 , 16 + 4 I 6 , 18 + 6 I 6 , 20 + 4 I 6 , 12 + I 6 , 24 = I 5 , 15 + 5 I 5 , 17 + 10 I 5 , 19 + 10 I 5 , 21 + 5 I 5 , 23 + I 5 , 25 = I 4 , 14 + 6 I 4 , 16 + 15 I 4 , 18 + 20 I 4 , 20 + 15 I 4 , 22 + 6 I 4 , 24 + I 4 , 26 = I 3 , 13 + 7 I 3 , 15 + 21 I 3 , 17 + 35 I 3 , 19 + 35 I 3 , 21 + 21 I 3 , 23 + 7 I 3 , 25 + I 3 , 27 = I 2 , 12 + 8 I 2 , 14 + 28 I 2 , 16 + 56 I 2 , 18 + 70 I 2 , 20 + 56 I 2 , 22 + 28 I 2 , 24 + 8 I 2 , 26 + I 2 , 28 = I 1 , 11 + 9 I 1 , 13 + 36 I 1 , 15 + 84 I 1 , 17 + 126 I 1 , 19 + 126 I 1 , 21 + 84 I 1 , 23 + 36 I 1 , 25 + 9 I 1 , 27 + I 1 , 29 = I 0 , 10 + 10 I 0 , 12 + 45 I 0 , 14 + 120 I 0 , 16 + 210 I 0 , 18 + 252 I 0 , 20 + 210 I 0 , 22 + 120 I 0 , 24 + 45 I 0 , 26 + 10 I 0 , 28 + I 0 , 30 \begin{aligned}I_{10,20}&=I_{9,19}+I_{9,21}\\&=I_{8,18}+2I_{8,20}+I_{8,22}\\&=I_{7,17}+3I_{7,19}+3I_{7,21}+I_{7,23}\\&= I_{6,16}+4I_{6,18}+6I_{6,20}+4I_{6,12}+I_{6,24}\\&=I_{5,15}+5I_{5,17}+10I_{5,19}+10I_{5,21}+5I_{5,23}+I_{5,25}\\&= I_{4,14}+6I_{4,16}+15I_{4,18}+20I_{4,20}+15I_{4,22}+6I_{4,24}+I_{4,26}\\&=I_{3,13}+7I_{3,15}+21I_{3,17}+35I_{3,19}+35I_{3,21}+21I_{3,23}+7I_{3,25}+I_{3,27}\\&=I_{2,12}+8I_{2,14}+28I_{2,16}+56I_{2,18}+70I_{2,20}+56I_{2,22}+28I_{2,24}+8I_{2,26}+I_{2,28}\\&=I_{1,11}+9I_{1,13}+36I_{1,15}+84I_{1,17}+126I_{1,19}+126I_{1,21}+84I_{1,23}+36I_{1,25}+9I_{1,27}+I_{1,29}\\&=I_{0,10}+10I_{0,12}+45I_{0,14}+120I_{0,16}+210I_{0,18}+252I_{0,20}+210I_{0,22}+120I_{0,24}+45I_{0,26}+10I_{0,28}+I_{0,30}\end{aligned}

Woah! All the coefficients are found in the Pascal's triangle ! Nice coincidence!

Pi Han Goh - 4 years, 5 months ago

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