Solutions - 0050-06

(1) Perform $u$-sub setting $u=\sin x$ and thus $du=\cos xdx$. Adjust the bounds as follows:

$$\begin{array}{cc}x& =0\gg \gg u=0\\ x& =\frac{\pi }{2}\gg \gg u=1\end{array}$$

Therefore:

$${\int }_0^{\pi /2}\frac{\cos x}{\sqrt{1+{\sin }^{2}x}}dx\gg \gg {\int }_0^1\frac{1}{\sqrt{1+u^2}}du$$

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(2) Notice $1+u^2$ pattern, so we should make use of the identity $1+{\tan }^2\theta ={\sec }^2\theta$.

Select $u=\tan \theta$ and thus $du={\sec }^2\theta d\theta$. Adjust bounds:

$$\begin{array}{cc}u& =0\gg \gg \theta =0\\ u& =1\gg \gg \theta =\frac{\pi }{4}\end{array}$$

Therefore:

$${\int }_0^1\frac{du}{\sqrt{1+u^2}}\gg \gg {\int }_0^{\pi /4}\frac{{\sec }^2\theta d\theta }{\sec \theta }\gg \gg {\int }_0^{\pi /4}\sec \theta d\theta$$

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(3) Integrate from memory or multiplying above and below by $\sec \theta +\tan \theta$:

$$\begin{array}{c}\gg \gg \ln |\sec \theta +\tan \theta |{|}_0^{\pi /4}\\ \\ \gg \gg \ln |\sec \frac{\pi }{4}+\tan \frac{\pi }{4}|-\ln |\sec 0+\tan 0|\\ \\ \gg \gg \ln |1+\sqrt{2}|-\ln 1\gg \gg \ln |1+\sqrt{2}|\end{array}$$