Question

Can some one solve this pls
`\int\limits^(1)/(√(2))_0 \frac{1}{\sqrt{1-x^(2) } }`

Answer 1

Explanation:

Hello, please consider the following.

`x(t)=sin(t)\\\\dx=cos(t)dt\\\\\text{For x = }(1)/(√(2))=(√(2))/(2) \text{ we have } t = (\pi)/(4)`

So, we can write.

`\displaystyle \int\limits^{(1)/(√(2))}_0 {(1)/(√(1-x^2))} \, dx =\int\limits^{(\pi)/(4)}_0 {(cos(t))/(√(1-sin^2(t)))} \, dt\\\\=\int\limits^{(\pi)/(4)}_0 {(cos(t))/(√(cos^2(t)))} \, dt\\\\=\int\limits^{(\pi)/(4)}_0 {(cos(t))/(cos(t))} \, dt \\\\=\int\limits^{(\pi)/(4)}_0 {1} \, dt\\\\=\large \boxed{\sf \bf (\pi)/(4)}`

Thank you

Answer 2

Answer:
`\bold{(\pi)/(4)}`

Explanation:

Note the following integral formula:
`\int\limits^a_b {(1)/(√(1-x^2))} \, dx =\sin^(-1)(x)\bigg|^a_b`

We can rationalize the denominator to get:
`(1)/(\sqrt2)\bigg((\sqrt2)/(\sqrt2)\bigg)=(\sqrt2)/(2)`

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`\int\limits^{(\sqrt2)/(2)}_0 {(1)/(√(1-x^2))} \, dx \\\\\\=\sin^(-1)(x)\bigg|^{(\sqrt2)/(2)}_0\\\\\\= \sin^(-1)\bigg((\sqrt2)/(2)\bigg)-\sin^(-1)(0)\\\\\\=(\pi)/(4)-0\pi\\\\\\=\large\boxed{(\pi)/(4)}`