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Answer:  $\frac{-1}{2 \sqrt{5}} \log \left[\frac{2 \sqrt{5}-\sqrt{5}-3}{2 \sqrt{5}+\sqrt{5}-3}\right]$

Hint: To solve this we convert  sin and cos into tan

Given:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{2 \cos x+4 \sin x} d x$

Solution:

Let

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{2 \cos x+4 \sin x} d x \\ & \end{aligned}

$I=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x+2 \sin x} d x$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{1-\tan ^{2} \frac{x}{2}+4 \tan ^{2} \frac{x}{2}} d x$

Let

\begin{aligned} &t=\tan ^{2} \frac{x}{2} \\ & \end{aligned}

$\frac{d t}{d x}=\sec ^{2} \frac{x}{2} \times \frac{1}{2}$

$2 d t=\sec ^{2} \frac{x}{2} d x$

Now

\begin{aligned} &\frac{1}{2} \times 2 \int_{0}^{1} \frac{d t}{1-t^{2}+4 t} \\ & \end{aligned}

$I=-\int_{0}^{1} \frac{d t}{t^{2}-4 t-1} \\$

$I=-\int_{0}^{1} \frac{d t}{t^{2}-4 t-1+2^{2}-2^{2}}$

\begin{aligned} &I=-\int_{0}^{1} \frac{d t}{(t-2)^{2}-(\sqrt{5})^{2}} \\ & \end{aligned}

$I=\frac{-1}{2 \sqrt{5}} \log \left|\frac{t-2-\sqrt{5}}{t-2+\sqrt{5}}\right|_{0}^{1} \\$

$I=\frac{-1}{2 \sqrt{5}} \log \left|\frac{-1-\sqrt{5}}{-1+\sqrt{5}}\right|-\log \left|\frac{-2-\sqrt{5}}{-2+\sqrt{5}}\right|$

\begin{aligned} &=\frac{-1}{2 \sqrt{5}} \log \left[\left|\frac{-1-\sqrt{5}}{-1+\sqrt{5}}\right| \times\left|\frac{-2-\sqrt{5}}{-2-\sqrt{5}}\right|\right] \\ & \end{aligned}

$=\frac{-1}{2 \sqrt{5}} \log \left(\frac{2-\sqrt{5}+2 \sqrt{5}-5}{2+\sqrt{5}-2 \sqrt{5}-\sqrt{5}}\right) \\$

$=\frac{-1}{2 \sqrt{5}} \log \left(\frac{2 \sqrt{5}-\sqrt{5}-3}{-2 \sqrt{5}+\sqrt{5}-3}\right)$

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