Question

Can some one solve this pls1%7D%7B%5Csqrt%7B1-x%5E%7B2%7D%20%7D%20%7D" id="TexFormula1" title="\int\limits^\frac{1}{\sqrt{2}}_0 \frac{1}{\sqrt{1-x^{2} } }" alt="\int\limits^\frac{1}{\sqrt{2}}_0 \frac{1}{\sqrt{1-x^{2} } }" align="absmiddle" class="latex-formula">

Answer 1

Answer:

Step-by-step explanation:

Hello, please consider the following.

$x(t)=sin(t)\\\\dx=cos(t)dt\\\\\text{For x = }\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2} \text{ we have } t = \dfrac{\pi}{4}$

So, we can write.

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

Thank you

Answer 2

Answer:  $\bold{\dfrac{\pi}{4}}$

Step-by-step explanation:

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

We can rationalize the denominator to get: $\dfrac{1}{\sqrt2}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)=\dfrac{\sqrt2}{2}$

*************************************************************************************

$\int\limits^{\frac{\sqrt2}{2}}_0 {\dfrac{1}{\sqrt{1-x^2}}} \, dx \\\\\\=\sin^{-1}(x)\bigg|^{\frac{\sqrt2}{2}}_0\\\\\\= \sin^{-1}\bigg(\dfrac{\sqrt2}{2}\bigg)-\sin^{-1}(0)\\\\\\=\dfrac{\pi}{4}-0\pi\\\\\\=\large\boxed{\dfrac{\pi}{4}}$