The integration ratio

Calculus Level 5

I n = ∫ 0 π e x ( sin ⁡ x ) n d x \large{I_{n}=\displaystyle \int^{\pi}_{0} e^{x}(\sin x)^{n} \, dx}

Define I n I_n as the integral above. If I 5 I 1 = a b \large{\dfrac{I_{5}}{I_{1}}=\dfrac{a}{b}} where a , b a,b are coprime integers, find a + b a+b .


The answer is 19.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

We try finding reduction formula. Consider

I n = ∫ 0 π ( sin ⁡ x ) n d e x = e x ( sin ⁡ x ) n − n ∫ 0 π e x ( sin ⁡ x ) n − 1 cos ⁡ x d x \displaystyle I_{n}=\int_{0}^{\pi} (\sin{x})^{n} d e^{x}= e^{x}(\sin{x})^{n}-n\int_{0}^{\pi} e^{x}(\sin{x})^{n-1} \cos{x} dx The first term is 0 provided that n is positive.

I n = − n ∫ 0 π ( sin ⁡ x ) n − 1 cos ⁡ x d e x = − n ( sin ⁡ x ) n − 1 cos ⁡ x e x + n ∫ 0 π e x [ ( n − 1 ) ( sin ⁡ x ) n − 2 cos ⁡ 2 x − ( sin ⁡ x ) n ] d x \displaystyle I_{n}=-n\int_{0}^{\pi} (\sin{x})^{n-1} \cos{x} d e^{x}=-n(\sin{x})^{n-1} \cos{x} e^{x}+n\int_{0}^{\pi} e^{x}[(n-1)(\sin{x})^{n-2} \cos^{2}{x}-(\sin{x})^{n}] dx The first term is again 0 provided n greater than 1. After simplifying the expression, we have

I n = n ( n − 1 ) n 2 + 1 I n − 2 \displaystyle I_{n}=\dfrac{n(n-1)}{n^{2}+1} I_{n-2}

So,

I 5 = 5 ⋅ 4 ⋅ 3 ⋅ 2 26 ⋅ 10 I 1 \displaystyle I_{5}=\dfrac{5 \cdot 4 \cdot 3 \cdot 2}{26 \cdot 10} I_{1} which implies a b = 6 / 13 \displaystyle \dfrac{a}{b}=6/13 and a + b = 19 \displaystyle a+b=\boxed{19}

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