Question

#PieDay
Prove the given integral.

$\displaystyle\int_{-1}^1 \dfrac{dx}{\sqrt{1-x^2}} = \pi$

Answer 1

We know that,

$\displaystyle\longrightarrow\int\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x$

So,

$\displaystyle\longrightarrow\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\left[\sin^{-1}x\right]_{-1}^1$

$\displaystyle\longrightarrow\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}(1)-\sin^{-1}(-1)$

$\displaystyle\longrightarrow\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)$

$\displaystyle\longrightarrow\underline{\underline{\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\pi}}$

OR

Let us be given to find,

$\displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}$

Put,

$\longrightarrow x=\sin\theta$

$\longrightarrow dx=\cos\theta\ d\theta$

$\longrightarrow x=-1\quad\implies\quad\theta=-\dfrac{\pi}{2}$

$\longrightarrow x=1\quad\implies\quad\theta=\dfrac{\pi}{2}$

Then the integral becomes,

$\displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos\theta\ d\theta}{\sqrt{1-\sin^2\theta}}$

$\displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos\theta\ d\theta}{\cos\theta}$

$\displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta$

$\displaystyle\longrightarrow I=\Big[\theta\Big]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\displaystyle\longrightarrow I=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)$

$\displaystyle\longrightarrow\underline{\underline{I=\pi}}$

Answer 2

Answer:

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