Wanna integrate by parts 10 times? (My ninth integral problem)

Calculus Level 5

∫ 0 π / 2 x 10 cos ⁡ x d x \int_0^{\pi/2} x^{10} \cos x \, dx

If the value of the integral above can be represented in the incredibly long form of

− a + b π 2 − c π 4 + d e π 6 − f g π 8 + π 10 h , -a + b \pi^{2} - c \pi^{4} + \dfrac{d}{e} \pi^{6} - \dfrac{f}{g} \pi^{8} + \dfrac{\pi^{10}}{h} ,

where a , b , c , d , e , f , g , h a,b,c,d,e,f,g,h are positive integers with gcd ⁡ ( d , e ) = gcd ⁡ ( f , g ) = 1 \gcd(d,e) = \gcd(f,g) = 1 , find a + b + c + d + e + f + g + h a+b+c+d+e+f+g+h .


The answer is 4093366.

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Hobart Pao by Hobart Pao , 23, USA

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2 solutions

Hamza A
Feb 8, 2016

Well,you can integrate by parts 10 times but you can notice a pattern.

The pattern is

∫ x n cos ⁡ ( x ) d x = sin ⁡ ( x ) ∑ i = 0 ⌊ n 2 ⌋ ( − 1 ) i n ! x m − 2 i ( n − 2 i ) ! + cos ⁡ ( x ) ∑ i = 0 ⌊ n − 1 2 ⌋ ( − 1 ) i n ! x n − 1 − 2 i ( n − 2 i − 1 ) ! \displaystyle\int { x^{ n }\cos { (x) } dx } =\sin { (x) } \displaystyle\sum _{ i=0 }^{ \left\lfloor \frac { n }{ 2 } \right\rfloor }{ (-1)^{ i } } \frac { n!x^{ m-2i } }{ (n-2i)! } +\cos (x)\displaystyle\sum _{ i=0 }^{ \left\lfloor \frac { n-1 }{ 2 } \right\rfloor }{ (-1)^{ i } } \frac { n!{ x }^{ n-1-2i } }{ (n-2i-1)! }

doing the rest we get the integral = π 10 1024 − 45 128 π 8 + 315 4 π 6 − 9450 π 4 + 453600 π 2 − 3628800 =\frac{\pi^{10}}{1024}-\frac{45}{128}\pi^8+\frac{315}{4}\pi^6-9450\pi^4+453600\pi^2-3628800

so answer = 1024 + 45 + 128 + 315 + 4 + 9450 + 453600 + 3628800 = 4093366 =1024+45+128+315+4+9450+453600+3628800=4093366

5 Helpful 1 Interesting 2 Brilliant 0 Confused

Or in simple terms, you did tabular integration :) nicely done

Hobart Pao - 5 years, 4 months ago

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Thanks!

and nice problem :)

Hamza A - 5 years, 4 months ago

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Thanks! :)

Hobart Pao - 5 years, 4 months ago

@Hobart Pao nice question. You should change the name of this problem. Because it gives a big hint.

Aditya Kumar - 5 years, 4 months ago

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You think it hints tabular too much?

Hobart Pao - 5 years, 4 months ago
Pi Han Goh
Apr 24, 2016

Tabular integration

u d v x 10 cos ⁡ x 10 x 9 − sin ⁡ x 90 x 8 − cos ⁡ x 720 x 7 sin ⁡ x 5040 x 6 cos ⁡ x 30240 x 5 − sin ⁡ x 151200 x 4 − cos ⁡ x 604800 x 3 − sin ⁡ x 181440 x 2 − cos ⁡ x 3628800 x sin ⁡ x 3628800 cos ⁡ x 0 − sin ⁡ x By tabular integration, we have ∫ x 10 cos ⁡ x d x = x 10 ( − sin ⁡ x ) − ( 10 x 9 ) ( − cos ⁡ x ) + ( 90 x 8 ) ( sin ⁡ x ) − ( 720 x 7 ) ( cos ⁡ x ) + ( 5040 x 6 ) ( − sin ⁡ x ) − ( 30240 x 5 ) ( − cos ⁡ x ) + ( 151200 x 4 ) ( − sin ⁡ x ) − ( 604800 x 3 ) ( − cos ⁡ x ) + ( 1814400 x 2 ) ( sin ⁡ x ) − ( 3628800 x ) ( cos ⁡ x ) + ( 3628800 x ) ( sin ⁡ x ) + C 0 0 Plugging in the limits gives ∫ 0 π / 2 x 10 cos ⁡ x d x = − 3628800 + 453600 π 2 − 9450 π 4 + 315 4 π 6 − 45 128 π 8 + 1 1024 π 10 \begin{array} { | c | c | } \hline \\ u & dv \\ \hline \\ x^{10} & \cos x \\ \hline \\ 10x^9 & -\sin x \\ \hline \\ 90x^8 & -\cos x \\ \hline \\ 720x^7 & \sin x \\ \hline \\ 5040x^6 & \cos x \\ \hline \\ 30240x^5 & -\sin x \\ \hline \\ 151200x^4 & -\cos x \\ \hline \\ 604800x^3 & -\sin x \\ \hline \\ 181440x^2 & -\cos x \\ \hline \\ 3628800x & \sin x \\ \hline \\ 3628800 & \cos x \\ \hline \\ 0 & -\sin x \\ \hline \end{array} \qquad \qquad \begin{aligned} &&\text{By tabular integration, we have} \\ && \int x^{10} \cos x \, dx \\ &=& x^{10} (-\sin x) - (10x^9)(-\cos x) + (90x^8)(\sin x) - (720x^7)(\cos x) \\ && + (5040x^6)(-\sin x) - (30240x^5)(-\cos x)+ (151200 x^4)(-\sin x) \\ && - (604800x^3)(-\cos x) + (1814400x^2) (\sin x) - (3628800x)(\cos x) \\ && + (3628800x)(\sin x) + C \\ \phantom0 \\ \phantom0 \\ && \text{Plugging in the limits gives} \\ &&\int_0^{\pi /2} x^{10} \cos x \, dx \\ &=& -3628800+453600 \pi^2-9450 \pi^4 +\dfrac{315}4 \pi^6 -\dfrac{45}{128} \pi^8+\dfrac1{1024} \pi^{10} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution! That's exactly what I did.

Hobart Pao - 5 years, 1 month ago

We can also write cosx as e^{ix} it suffices to compute real part but it's literally the same as above

<> <> - 3 years, 10 months ago

Though your final answer is correct, your table is wrong (and your mistakes there spill over into your indefinite integral). First, 181440 should be 1814400. Second, your signs are wrong; for instance, x^10 (-sin x) would give you -(1/1024) pi^10 (not +).

Linden Yuan - 10 months, 3 weeks ago

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Whoops. You're right. I've fixed the first part already.

Pi Han Goh - 10 months, 3 weeks ago

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