A simple integral

Calculus Level 3

0 2 π a ( 1 cos t ) d t \large \int_0^{2 \pi} \sqrt{a(1 - \cos t)} \ dt

Find the value of the integral above, where a > 0 a > 0 .

2 a \sqrt{2a} 4 2 a 4 \sqrt{2a} 2 2 a 2 \sqrt{2a} 2 a 2 \sqrt{a}

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2 solutions

Chew-Seong Cheong
Apr 30, 2017

I = 0 2 π a ( 1 cos t ) d t = 0 2 π a ( 1 ( 1 2 sin 2 t 2 ) ) d t = 0 2 π 2 a sin t 2 d t = 2 2 a cos t 2 0 2 π = 4 2 a \begin{aligned} I & = \int_0^{2\pi} \sqrt{a(1-\color{#3D99F6}\cos t)} \ dt \\ & = \int_0^{2\pi} \sqrt{a\left(1-\color{#3D99F6}\left(1-2\sin^2 \frac t2\right)\right)} \ dt \\ & = \int_0^{2\pi} \sqrt{2a} \sin \frac t2 \ dt \\ & = -2 \sqrt{2a} \cos \frac t2 \bigg|_0^{2\pi} \\ & = \boxed{4\sqrt{2a}} \end{aligned}

Rocco Dalto
Jun 1, 2017

0 2 π a ( 1 c o s t ) d t = 2 2 a 0 2 π s i n ( t 2 ) 1 2 d t = \int_{0}^{2\pi} \sqrt{a(1 - cost)} dt = 2 \sqrt{2a} \int_{0}^{2\pi} sin(\dfrac{t}{2}) * \dfrac{1}{2} dt =

2 2 a c o s ( t 2 ) 0 2 π = 4 2 a -2 \sqrt{2a} * cos(\dfrac{t}{2})|_{0}^{2\pi} = \boxed{4 \sqrt{2a}}

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