can some one solve this pls[tex]\int\limits^\frac{1}{\sqrt{2}}_0 \frac{1}{\sqrt{1-x^{2} } }[/tex]

[tex]\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}}[/tex]

Thank you

tramserran tramserran · Answer 2 · 2020-08-15T11:22:44+02:00

Answer: [tex]\bold{\dfrac{\pi}{4}}[/tex]

Step-by-step explanation:

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

We can rationalize the denominator to get: [tex]\dfrac{1}{\sqrt2}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)=\dfrac{\sqrt2}{2}[/tex]

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[tex]\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}}[/tex]