Question

$\large{\mathcal{\colorbox{gold}{\red{Prove the given integral.️}}}}$

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

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Answer 1

Answer:

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Step-by-step explanation:

$We \: \: know \: \: that \\ \\ \displaystyle\longrightarrow\int\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x \\ \\ So \\ \\ \displaystyle\longrightarrow\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\left[\sin^{-1}x\right]_{-1}^1 \\ \\ \displaystyle\longrightarrow\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}(1)-\sin^{-1}(-1)\\ \\ \displaystyle\longrightarrow\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right) \\ \\ \displaystyle\longrightarrow\underline{\underline{\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}}=\pi}} \\ \\ OR \\ \\ Let \: \: us \: \: be \: \: given \: \: to \: \: find, \\ \\\displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{dx}{\sqrt{1-x^2}} \\ \\ Put \\ \\ \longrightarrow x=\sin\theta \\ \\ \longrightarrow dx=\cos\theta\ d\theta \\ \\ \longrightarrow x=-1\quad\implies\quad\theta=-\dfrac{\pi}{2} \\ \\ \longrightarrow x=1\quad\implies\quad\theta=\dfrac{\pi}{2} \\ \\ Then \: \: the \: \: integral \: \: becomes, \\ \\ \displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos\theta\ d\theta}{\sqrt{1-\sin^2\theta}} \\ \\ \displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos\theta\ d\theta}{\cos\theta} \\ \\ \displaystyle\longrightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta \\ \\ \displaystyle\longrightarrow I=\Big[\theta\Big]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ \\ \displaystyle\longrightarrow I=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right) \\ \\ \displaystyle\longrightarrow\underline{\underline{I=\pi}}$

Answer 2

Answer:

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Step-by-step explanation:

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