Cosine and few friends

Calculus Level 4

2016 π 2015 π ( cos ( x ) + cos ( x ) + cos ( x ) ) d x = ? \large \displaystyle \int_{-2016 \pi}^{2015 \pi} (\cos(x) + |\cos(x)|+ \cos(|x|)) dx = ?

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Graph credit: Math is Fun


The answer is 8062.

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2 solutions

2016 π 2015 π ( cos ( x ) + cos ( x ) + cos ( x ) ) d x = 2016 π 2015 π cos ( x ) d x + 2016 π 2015 π cos ( x ) d x + 2016 π 2015 π cos ( x ) d x = [ sin ( x ) ] 2016 π 2015 π + 2015.5 π 2015.5 π cos ( x ) d x + 0 2015 π cos ( x ) d x + 0 2016 π cos ( x ) d x = [ sin ( x ) ] 0 2015 π + [ sin ( x ) ] 0 2016 π + 8062 0 1 2 π cos ( x ) d x = 8062 \int_{-2016 \pi}^{2015 \pi} (\cos(x) + |\cos(x)|+ \cos(|x|)) dx\\=\int_{-2016 \pi}^{2015 \pi}\cos(x)dx+\int_{-2016 \pi}^{2015 \pi}|\cos(x)|dx+\int_{-2016 \pi}^{2015 \pi}\cos(|x|)dx\\=\left[\sin(x)\right]^{2015\pi}_{-2016\pi}+\int_{-2015.5 \pi}^{2015.5 \pi}|\cos(x)|dx+\int_{0}^{2015 \pi}\cos(x)dx+\int_{0}^{2016 \pi}\cos(x)dx\\=\left[\sin(x)\right]^{2015\pi}_{0}+\left[\sin(x)\right]^{2016\pi}_{0}+8062\int_{0}^{\frac{1}{2} \pi} \cos(x)dx\\=8062

Chew-Seong Cheong
Feb 19, 2018

Let I = 2016 π 2015 π cos x + cos x + cos x d x = I 1 + I 2 + I 3 I = \displaystyle \int_{-2016 \pi}^{2015 \pi} \cos x + |\cos x| + \cos |x| \ dx = I_1 + I_2 + I_3 respectively.

We note that c o s x cos x is an even function with a period or cycle of 2 π 2\pi and that I c = a b cos x d x = 0 I_c = \displaystyle \int_a^b \cos x \ dx = 0 , when b a = n π b-a=n\pi , where n n is an integer. Then we have:

I 1 = 2016 π 2015 π cos x d x = 0 Since b a = 4031 π \begin{aligned} I_1 & = \int_{-2016 \pi}^{2015 \pi} \cos x \ dx = 0 & \small \color{#3D99F6} \text{Since }b-a = 4031\pi \end{aligned}

I 2 = 2016 π 2015 π cos x d x All half-cycles are positive. = ( 2015 ( 2016 ) ) × 2 0 π 2 cos x d x = 8062 sin x 0 π 2 = 8062 \begin{aligned} I_2 & = \int_{-2016 \pi}^{2015 \pi} |\cos x| \ dx & \small \color{#3D99F6} \text{All half-cycles are positive.} \\ & = \left(2015-(-2016)\right) \times 2\int_0^\frac \pi 2 \cos x \ dx \\ & = 8062 \sin x \bigg|_0^\frac \pi 2 \\ & = 8062 \end{aligned}

I 1 = 2016 π 2015 π cos x d x = 0 Note that cos x = cos x as both are even = 2016 π 2015 π cos x d x = 0 Since b a = 4031 π \begin{aligned} I_1 & = \int_{-2016 \pi}^{2015 \pi} \cos |x| \ dx = 0 & \small \color{#3D99F6} \text{Note that }\cos |x| = \cos x \text{ as both are even} \\ & = \int_{-2016 \pi}^{2015 \pi} \cos x \ dx = 0 & \small \color{#3D99F6} \text{Since }b-a = 4031\pi \end{aligned}

Therefore, I = I 1 + I 2 + I 3 = 0 + 8062 + 0 = 8062 I = I_1+I_2+I_3 = 0+8062+0 = \boxed{8062} .

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