A calculus problem by Rahil Sehgal

Calculus Level 2

∫ 0 π / 4 cos ⁡ 7 2 x d x \large \int_{0}^{\pi/4} \cos^7 2x \ dx

If the value of the integral above is in the form of a b \dfrac {a}{b} , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 43.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Chew-Seong Cheong
Apr 29, 2017

I = ∫ 0 π 4 cos ⁡ 7 2 x d x Let u = 2 x , d u = 2 d x = 1 2 ∫ 0 π 2 cos ⁡ 7 u d u = 1 2 ∫ 0 π 2 ( 1 − sin ⁡ 2 u ) 3 cos ⁡ u d u Let t = sin ⁡ u , d t = cos ⁡ u d u = 1 2 ∫ 0 1 ( 1 − t 2 ) 3 d t = 1 2 ∫ 0 1 ( 1 − 3 t 2 + 3 t 4 − t 6 ) d t = 1 2 [ t − t 3 + 3 t 5 5 − t 7 7 ] 0 1 = 1 2 ( 1 − 1 + 3 5 − 1 7 ) = 8 35 \begin{aligned} I & = \int_0^\frac \pi 4 \cos^7 2x \ dx & \small \color{#3D99F6} \text{Let }u = 2x, \ du = 2dx \\ & = \frac 12 \int_0^\frac \pi 2 \cos^7 u \ du \\ & = \frac 12 \int_0^\frac \pi 2 \left(1-\sin^2 u \right)^3\cos u \ du & \small \color{#3D99F6} \text{Let }t = \sin u, \ dt = \cos u du \\ & = \frac 12 \int_0^1 \left(1-t^2\right)^3 \ dt \\ & = \frac 12 \int_0^1 \left(1-3t^2 + 3t^4-t^6\right) \ dt \\ & = \frac 12 \left[t- t^3 + \frac {3t^5}5 - \frac {t^7}7 \right]_0^1 \\ & = \frac 12 \left(1- 1 + \frac 35 - \frac 17 \right) \\ & = \boxed{\dfrac 8{35}} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Good solution. After a long while did I see that you uploaded a solution with an integral involved like ∫ 0 π / 2 2 sin ⁡ 2 m − 1 θ cos ⁡ 2 n − 1 θ d θ \displaystyle \int_0^{{\pi}/{2}} 2 \sin^{2m-1} \theta \cos^{2n-1} \theta \,d \theta and didn't replace it with beta function followed by gamma function. (+1) for this elementary approach sir.

Tapas Mazumdar - 4 years, 1 month ago

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