Question - Solving an Arcsecant Integral
Solution:
The integral in the image is:$$\int_0^{\frac{1}{2}} \frac{dx}{\sqrt{1 - x^2}}$$This integral is the integral of the arcsecant derivative. We can use the substitution $$ x = \sin(\theta) $$ to solve it:$$dx = \cos(\theta) d\theta$$When $$ x = 0 $$, $$ \sin(\theta) = 0 $$, so $$ \theta = 0 $$.When $$ x = \frac{1}{2} $$, $$ \sin(\theta) = \frac{1}{2} $$, which means $$ \theta = \frac{\pi}{6} $$ since it lies in the first quadrant.Substituting $$ x = \sin(\theta) $$ into the integral gives us:$$\int_0^{\frac{\pi}{6}} \frac{\cos(\theta)}{\sqrt{1 - \sin^2(\theta)}} d\theta$$Now, $$ \sqrt{1 - \sin^2(\theta)} = \sqrt{\cos^2(\theta)} = \cos(\theta) $$So the integral simplifies to:$$\int_0^{\frac{\pi}{6}} d\theta$$This is just the integral of $$1$$ with respect to $$ \theta $$ from $$0$$ to $$\frac{\pi}{6}$$, so it equals $$ \theta $$ evaluated between these two limits:$$\theta \Big|_0^{\frac{\pi}{6}} = \frac{\pi}{6} - 0 = \frac{\pi}{6}$$Therefore, the integral equals $$ \frac{\pi}{6} $$.
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