Question

#PieDay

Prove that.
$\pi = 4\int_0^1 \sqrt{1-x^2} dx$

Answer 1

$\bigstar\:\rm{We\:need\:to\:prove\:that\::\pi=4 \int_{0}^{1} \sqrt{1-x^{2}} d x}\longrightarrow\rm{4 \int_{0}^{1} \sqrt{1-x^{2}} d x=\pi}$

$\longrightarrow\rm{4 \int_{0}^{1} \sqrt{1-x^{2}} d x=4\left[\frac{1}{2}\left(\sin ^{-1} x+x \sqrt{1-x^{2}}\right)\right]_{0}^{1}}$

$\rm{...\:Since \int \sqrt{1-x^{2}} d x=\frac{1}{2}\left[\sin ^{-1} x+x \sqrt{1-x^{2}}\right]+C}$

$\textsf{Find the calculation of \rm{\int \sqrt{1-x^{2}} d x} at bottom of the answer}$

$\longrightarrow\rm{4 \int_{0}^{1} \sqrt{1-x^{2}} d x=2\left[\left(\sin ^{-1} 1+(1) \sqrt{1-(1)^{2}}\right)-\left(\sin ^{-1} 0+(0) \sqrt{1-(0)^{2}}\right)\right]}$

$\longrightarrow\rm{4 \int_{0}^{1} \sqrt{1-x^{2}} d x=2[\pi/2]=\pi\quad...\:Hence\:Proved!}$

$\rule{315}{2}$

$\bigstar\:\rm{Calculation\: of\: \int \sqrt{1-x^{2}} d x}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=\int \sqrt{1-\sin ^{2} \theta} \cos \theta d \theta}$

$\rm{...\:Putting\:x=\sin\theta\implies1=\cos \theta \cdot d \theta/d x\implies d x=\cos \theta \cdot d \theta}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=\int \sqrt{\cos ^{2} \theta} \cos \theta d \theta}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=\int \cos \theta\times \cos \theta\: d \theta=\int \cos^2 \theta d \theta}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=\int (1+\cos 2 \theta)/2 \:d \theta\quad...\:Since\:\cos 2 x=2 \cos ^{2} x-1}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=1/2 \int(1+\cos 2 \theta) d \theta=1/2\left[\theta+(\sin 2 \theta)/2\right]+C}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=1/2\left[\sin ^{-1} x+\frac{\sin \left(2 \sin ^{-1} x\right)}{2}\right]+C\quad\because\:x=\sin \theta \Rightarrow \theta=\sin ^{-1} x}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=1/2\left[\sin ^{-1} x+\frac{\sin \left(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\right)}{2}\right]+C\:\because 2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)}$

$\longrightarrow\rm{\int \sqrt{1-x^{2}} d x=1/2\left[\sin ^{-1} x+\frac{2 x \sqrt{1-x^{2}}}{2}\right]+C=1/2\left[\sin ^{-1} x+x \sqrt{1-x^{2}} \mid+C\right.}$

Answer 2

$\bf \: Prove \: that : \: \pi = 4\int_0^1 \sqrt{1-x^2} dx$

$\large\underline{\bold{Solution-}}$

Consider, RHS

$\tt \:I = 4\int_0^1 \sqrt{1-x^2} dx - - - (1)$

To evaluate this integral, we use method of Substitution.

So,

$\tt \: Put \: x = \: siny \: - - - (2)$

$\therefore \: \: \: \tt \: \dfrac{dx}{dy} = cosy$

$\tt \: dx = cosy \: dy \: - - - (3)$

Now limits changes to

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf x = siny \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf \dfrac{\pi}{2} \end{array}} \\ \end{gathered}$

Now, Substitute all these values in equation (1), we get

$\tt \: I = 4\int_0^ \frac{\pi}{2} \sqrt{1-(siny)^2} \: cosy \: dy$

$\tt \: I = 4\int_0^ \frac{\pi}{2} \sqrt{(cosy)^2} \: cosy \: dy$

$\tt \: I = 4\int_0^ \frac{\pi}{2} \: {cos}^{2}y \: dy$

$\tt \: I = 2\int_0^ \frac{\pi}{2} \: 2{cos}^{2}y \: dy$

$\tt \: I = 2\int_0^ \frac{\pi}{2} (1 + cos2y) \: dy$

$\tt \: I =2 \bigg(y + \dfrac{sin2y}{2} \bigg)_0^ \dfrac{\pi}{2}$

$\tt \: I = 2\bigg(\dfrac{\pi}{2} + \dfrac{sin\pi}{2} \bigg)$

$\tt \: I = 2\bigg( \dfrac{\pi}{2} - 0\bigg)$

$\tt \: I = \pi$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \tt{ \int \: sinx \: dx = - \: cosx \: + \: c}}$

$\boxed{ \tt{ \int \: cosx \: dx = sinx \: + \: c}}$

$\boxed{ \tt{ \int \: {sec}^{2}x \: dx = tan \: x \: + \: c}}$

$\boxed{ \tt{ \int \: {cot}^{2} x \: dx = - \: cosecx \: + \: c}}$