Pre-interview_Test_Question_16

Screenshot 2020-07-05 at 23.21.17

Let x = \sin \theta then \mathrm dx = \cos \theta \mathrm d\theta. Note also that when \theta = 0, x = 0 and when \theta = \dfrac \pi 2, x = 1. Then:

\begin{align*}\int_0^1 \frac {\mathrm dx} {x + \sqrt {1-x^2}} & = \int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \sqrt {1 - \sin^2 \theta}} \mathrm d\theta \\ & = \int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \sqrt {\cos^2 \theta}} \mathrm d\theta\end{align*}

Since \cos \theta \ge 0 for 0 \le \theta \le \dfrac \pi 2 we have:

\displaystyle \int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \sqrt {\cos^2 \theta}} \mathrm d\theta = \int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \cos \theta} \mathrm d\theta

Applying a substitution of \displaystyle \theta = \frac \pi 2 - \phi we have:

\begin{align*}\int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \cos \theta} \mathrm d\theta & = -\int_{\pi/2}^0 \frac {\cos \left(\frac \pi 2 - \phi\right)} {\sin \left(\frac \pi 2 - \phi\right) + \cos \left(\frac \pi 2 - \phi\right)} \mathrm d\phi \\ & = \int_0^{\pi/2} \frac {\sin \phi} {\sin \phi + \cos \phi} \mathrm d\phi\end{align*}

Adding these gives:

\displaystyle 2 \int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \cos \theta} \mathrm d\theta = \int_0^{\pi/2} \frac {\sin \theta + \cos \theta} {\sin \theta + \cos \theta} \mathrm d\theta = \int_0^{\pi/2} \mathrm d\theta = \frac \pi 2

giving:

\displaystyle \int_0^1 \frac {\mathrm dx} {x + \sqrt {1-x^2}} = \int_0^{\pi/2} \frac {\cos \theta} {\sin \theta + \cos \theta} \mathrm d\theta = \frac \pi 4

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This is unreal. @_gcx - how did you intuit x= \sin \theta? Is it the \sqrt{1-x^2}?

Yeah pretty much and the substitution was from my STEP mental toolkit.

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Battle-hardened