Prove the given integral.

1.	Prove the given integral.
Answer» $\huge\bold{\mathtt{\colorbox{yellow}{\blue{Questioñ↓}}}}$ Prove the GIVEN integral. $\displaystyle\int_{-1}^1 \dfrac{dx}{\sqrt{1-x^2}} = \pi$ $\huge\bold{\mathtt{\colorbox{gold}{\red{Añswer↓}}}}$ $\Large\red{\fbox{\displaystyle\underline{\underline{I=π}}}}$ $\huge\bold{\mathtt{\colorbox{orange}{\red{Solutioñ↓}}}}$ 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, $\displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos\theta\ d\theta}{\sqrt{1-\sin^2\theta}}$ $\displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos\theta\ d\theta}{\cos\theta}$ $\displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta$ $\displaystyle\longrightarrow I=\Big[\theta\Big]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[θ]$ $\displaystyle\longrightarrow I=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)$ $\Large\red{\fbox{\displaystyle\underline{\underline{Answer=I=π}}}}$

Answer»

$\huge\bold{\mathtt{\colorbox{yellow}{\blue{Questioñ↓}}}}$