Summation of integrals!

Calculus Level 3

For a k = ∫ 0 π 2 sin ⁡ 2 x ( 1 − sin ⁡ x ) k − 1 2 d x a_{k} =\displaystyle \int_{0}^{\frac{\pi}{2}} \sin 2x {(1 - \sin x)}^{\frac{k-1}{2}} \ dx , where k ∈ N k \in \mathbb N , ∑ k = 3 ∞ ( k + 1 ) ( a k − a k + 1 ) \displaystyle \sum_{k=3}^{\infty} (k+1)(a_{k} - a_{k+1}) can be expressed as a b \dfrac{a}{b} , where a a and b b are coprime positive integers. Find a b ab .


The answer is 70.

Jatin Yadav by jatin yadav , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Jatin Yadav
Dec 27, 2013

a k = 2 ∫ 0 π 2 sin ⁡ x ( 1 − sin ⁡ x ) k − 1 2 cos ⁡ x dx a_{k} = \displaystyle 2 \int_{0}^{\frac{\pi}{2}} \sin x (1 - \sin x)^{\frac{k-1}{2}} \cos x \text{dx}

Substituting 1 − sin ⁡ x = t 1 - \sin x = t yields

a k = 2 ∫ 0 1 t k − 1 2 ( 1 − t ) dt a_{k} = 2\displaystyle \int_{0}^{1} t^{\frac{k-1}{2}}(1-t) \text{dt}

= 2 ( ∫ 0 1 t k − 1 2 dt + ∫ 0 1 t t + 1 2 dt ) 2 \bigg( \displaystyle \int_{0}^{1} t^{\frac{k-1}{2}} \text{dt} + \int_{0}^{1} t^{\frac{t+1}{2}} \text{dt}\bigg)

= 4 ( 1 k + 1 − 1 k + 3 ) 4 \bigg( \frac{1}{k+1} - \frac{1}{k+3} \bigg)

Now,

∑ k = 3 ∞ ( k + 1 ) ( a k − a k + 1 ) \displaystyle \sum_{k=3}^{\infty} (k+1)(a_{k} - a_{k+1})

= 4 ∑ k = 3 ∞ ( k + 1 k + 1 − k + 1 k + 3 − k + 1 k + 2 + k + 1 k + 4 ) \displaystyle 4\sum_{k=3}^{\infty} \bigg(\frac{k+1}{k+1} - \frac{k+1}{k+3} - \frac{k+1}{k+2} + \frac{k+1}{k+4}\bigg)

= 4 ∑ k = 3 ∞ ( 1 − ( 1 − 2 k + 3 ) − ( 1 − 1 k + 2 ) + ( 1 − 3 k + 4 ) ) \displaystyle 4 \sum_{k=3}^{\infty} \Bigg(1 - \bigg(1 - \frac{2}{k+3}\bigg) - \bigg(1 - \frac{1}{k+2}\bigg) + \bigg(1 - \frac{3}{k+4} \bigg)\Bigg)

= 4 ( ∑ k = 3 ∞ ( 1 k + 2 − 1 k + 3 ) + 3 ∑ k = 3 ∞ ( 1 k + 3 − 1 k + 4 ) ) \displaystyle 4 \Bigg( \sum_{k=3}^{ \infty}\bigg( \frac{1}{k+2} - \frac{1}{k+3}\bigg) + 3 \sum_{k=3}^{\infty} \bigg(\frac{1}{k+3} - \frac{1}{k+4} \bigg) \Bigg)

= 4 ( 1 5 + 3 1 6 ) \displaystyle 4 \bigg( \frac{1}{5} + 3 \frac{1}{6} \bigg)

= 14 5 \boxed{\frac{14}{5}}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

but dt=- cos x dx so there should be an extra minus sign

Ashar Tafhim - 7 years, 5 months ago

Log in to reply

note: the limits for t were 1 to 0, and there was (-dt), but i reversed the limits and cancelled the - sign.

jatin yadav - 7 years, 5 months ago
Chew-Seong Cheong
Aug 27, 2018

Similar solution with @jatin yadav 's

a k = ∫ 0 1 sin ⁡ 2 x ( 1 − sin ⁡ x ) k − 1 2 d x = ∫ 0 1 2 sin ⁡ x cos ⁡ x ( 1 − sin ⁡ x ) k − 1 2 d x Let u = sin ⁡ x ⟹ d u = cos ⁡ x d x = 2 ∫ 0 1 u ( 1 − u ) k − 1 2 d u By ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = 2 ∫ 0 1 ( 1 − u ) u k − 1 2 d u = 2 ∫ 0 1 ( u k − 1 2 − u k + 1 2 ) d u = 2 [ 2 u k + 1 2 k + 1 − 2 u k + 3 2 k + 3 ] 0 1 = 4 ( 1 k + 1 − 1 k + 3 ) \begin{aligned} a_k & = \int_0^1 \sin 2x(1-\sin x)^{\frac {k-1}2} dx \\ & = \int_0^1 2 \sin x \cos x(1-\sin x)^{\frac {k-1}2} dx & \small \color{#3D99F6} \text{Let }u = \sin x \implies du = \cos x \ dx \\ & = 2 \int_0^1 u(1-u)^{\frac {k-1}2} du & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = 2 \int_0^1 (1-u)u^{\frac {k-1}2} du \\ & = 2 \int_0^1 \left(u^{\frac {k-1}2} - u^{\frac {k+1}2} \right) du \\ & = 2 \left[\frac {2u^{\frac {k+1}2}}{k+1} - \frac {2u^{\frac {k+3}2}}{k+3} \right]_0^1 \\ & = 4 \left(\frac 1{k+1} - \frac 1{k+3} \right) \end{aligned}

Now consider,

S = ∑ k = 3 ∞ ( k + 1 ) ( a k − a k + 1 ) = 4 a 3 − 4 a 4 + 5 a 4 − 5 a 5 + 6 a 5 − 6 a 6 + ⋯ = 4 a 3 + a 4 + a 5 + a 6 + ⋯ = 3 a 3 + ∑ k = 3 ∞ a k = 12 ( 1 4 − 1 6 ) + 4 ∑ k = 3 ∞ ( 1 k + 1 − 1 k + 3 ) = 3 − 2 + 4 ( 1 4 + 1 5 ) = 14 5 \begin{aligned} S & = \sum_{k=3}^\infty (k+1) \left(a_k - a_{k+1}\right) \\ & = 4a_3 - 4a_4 + 5a_4 - 5a_5 + 6a_5 - 6a_6 + \cdots \\ & = 4a_3 + a_4 + a_5 + a_6 + \cdots \\ & = 3a_3 + \sum_{k=3}^\infty a_k \\ & = 12\left(\frac 14 - \frac 16\right) + 4\sum_{k=3}^\infty \left(\frac 1{k+1} - \frac 1{k+3} \right) \\ & = 3 - 2 + 4\left(\frac 14 + \frac 15\right) \\ & = \frac {14}5 \end{aligned}

Therefore, a b = 14 × 5 = 70 ab = 14\times 5 = \boxed{70} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...