Calculus And Computer Science?

Calculus Level 4

∫ 0 π sin ⁡ 20 x d x \large \displaystyle \int_{0}^{\pi} \sin^{20}x \, dx

The integral above has a closed form. Find this closed form.

Give your answer to 3 decimal places.

Bonus : Compute ∫ 0 π sin ⁡ 80 x d x \displaystyle \int_{0}^{\pi} \sin^{80}x \, dx .

Try a harder version here: Calculus and Computer Science Again?


The answer is 0.55354.

Kunal Verma by Kunal Verma , 22, India

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3 solutions

Rishabh Jain
May 7, 2016

I n = ∫ 0 π sin ⁡ n x d x = ∫ 0 π sin ⁡ n − 1 x sin ⁡ x d x \large I_n=\displaystyle\int_0^{\pi}\sin^n x\mathrm{d}x\\=\large\displaystyle\int_0^{\pi}\sin^{n-1}x\color{#D61F06}{ \sin x}\mathrm{d}x Using Integration by Parts:- I n = − sin ⁡ n − 1 x cos ⁡ x ∣ 0 π ⏟ 0 + ( n − 1 ) ∫ 0 π sin ⁡ n − 2 x cos ⁡ 2 x ⏟ 1 − sin ⁡ 2 x d x I_n=\underbrace{-\sin^{n-1} x\cos x|_0^{\pi}}_{\large 0}+(n-1)\displaystyle\int_0^{\pi}\sin^{n-2}x\underbrace{\cos^2x}_{\color{teal}{1-\sin^2 x}}\mathrm{d}x

⟹ I n = ( n − 1 ) ( I n − 2 − I n ) \implies I_n=(n-1)(I_{n-2}-I_{n}) ⟹ I n = n − 1 n ( I n − 2 ) \large \implies I_n=\dfrac{n-1}{n}(I_{n-2})

Hence,

I 20 = ( 19 20 ) × ( 17 18 ) × ( 15 16 ) × ⋯ × ( 1 2 ) × I 0 ⏟ π I_{20}=\left(\dfrac{19}{20}\right)\times\left(\dfrac{17}{18}\right)\times\left(\dfrac{15}{16}\right)\times\cdots\times\left(\dfrac{1}{2}\right)\times \underbrace{I_0}_{\large \pi}

= 46189 π 262144 ≈ 0.55354 \large =\dfrac{46189\pi}{262144}\approx \boxed{0.55354}

Generalised form:- I n = ( n − 1 ) ! ! π n ! ! \Large I_n=\dfrac{(n-1)!!\pi}{n!!}

(where ! ! !! is double factorial notation).

16 Helpful 0 Interesting 1 Brilliant 0 Confused

Adding to your solution, the recursive function to evaluate I n I_{n} in C++: 

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#include <iostream>

using namespace std;

float recint(int n)
{
    float ans;
    if(n==0)
        return 3.1415;
    ans=(n-1)*recint(n-2)/n;
    return ans;
}
int main()
{
    cout << recint(20) << " " << recint(80);
    return 0;
}

Output :

0.553523 0.553523 and 0.279367 0.279367 respectively for I 20 I_{20} and I 80 I_{80} in 0.686s.

Kunal Verma - 5 years, 1 month ago

Bonus : Calculate I 80 I_{80}

Kunal Verma - 5 years, 1 month ago

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I 80 ≈ 0.279375 ( = 79 ! ! π 80 ! ! ) . I_{80}\approx 0.279375~~ (=\dfrac{79!!\pi}{80!!}).

Rishabh Jain - 5 years, 1 month ago

Shall I delete it ... ?? (As it does not involve programming obviously)..

Rishabh Jain - 5 years, 1 month ago

Log in to reply

No. Programming is appreciated but not necessary. You gave a valid solution and that's what matters I think.

Kunal Verma - 5 years, 1 month ago

about the generalised form is that I n { I }_{ n } is not true when n n is odd.

Joel Yip - 4 years, 6 months ago

I = ∫ 0 π sin ⁡ 20 x d x = ∫ 0 π ( 2 sin ⁡ ( x 2 ) cos ⁡ ( x 2 ) ) 20 d x Let θ = x 2 ⟹ d θ = 1 2 d x = 2 21 ∫ 0 π 2 sin ⁡ 20 θ cos ⁡ 20 θ d θ = 2 20 B ( 21 2 , 21 2 ) B ( m , n ) is Beta function = 2 20 Γ 2 ( 21 2 ) Γ ( 21 ) Γ ( n ) is Gamma function = 2 20 ⋅ ( 19 ! ! 2 10 ) 2 π 20 ! = ( 19 ! ! ) 2 π 20 ! = 19 ! ! π 2 10 10 ! ≈ 0.554 \begin{aligned} I & = \int_0^\pi \sin^{20} x \, dx \\ & = \int_0^\pi \left(2\sin \left(\frac{x}{2} \right) \cos \left(\frac{x}{2} \right) \right)^{20} \, dx \quad \quad \small \color{#3D99F6}{\text{Let }\theta = \frac{x}{2} \implies d\theta = \frac{1}{2} \ dx} \\ & = 2^{21} \int_0^{\frac{\pi}{2}} \sin^{20} \theta \cos^{20} \theta \ d\theta \\ & = 2^{20} B \left(\frac{21}{2}, \frac{21}{2} \right) \quad \quad \small \color{#3D99F6}{B(m,n) \text{ is Beta function}} \\ & = \frac{2^{20} \Gamma^2 \left(\frac{21}{2} \right)}{\Gamma (21)} \quad \quad \small \color{#3D99F6}{\Gamma (n) \text{ is Gamma function}} \\ & = \frac{2^{20}\cdot \left(\frac{19!!}{2^{10}}\right)^2\pi}{20!} \\ & = \frac{(19!!)^2\pi}{20!} \\ & = \frac{19!!\pi}{2^{10}10!} \approx \boxed{0.554} \end{aligned}

General form: I n = ∫ 0 1 sin ⁡ n ( x ) d x = ( n − 1 ) ! ! π n ! ! \ \ I_n = \displaystyle \int_0^1 \sin^n (x) \ dx = \frac{(n-1)!!\pi}{n!!}

10 Helpful 1 Interesting 1 Brilliant 0 Confused

Firstly, for the general form, the upper limit is pi and the generalised form is not true when n n is odd.

Joel Yip - 4 years, 6 months ago

Since you wanted a Computer Science solution: 

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import math
def f(x):
    return (math.sin(x))**20
i=0
s=0
while(i<3.14159265):
    s += abs(f(i))*0.000001
    i += 0.000001
print(s)

Bonus: 

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import math
def f(x):
    return (math.sin(x))**80
i=0
s=0
while(i<3.14159265):
    s += abs(f(i))*0.000001
    i += 0.000001
print(s)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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