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Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 60

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Answer:   \frac{1}{2 \sqrt{5}} \log \left|\frac{1-\frac{1}{2}-\frac{\sqrt{5}}{2}}{1-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|-\log \left|\frac{1+\sqrt{5}}{1-\sqrt{5}}\right|

Hint: To this equation we convert cos and sin in terms of tan

Given:  \int_{0}^{\frac{\pi}{2}} \frac{d x}{4 \cos x+2 \sin x}

Solution:  I= \int_{0}^{\frac{\pi}{2}} \frac{d x}{4 \cos x+2 \sin x}

\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]

\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(\frac{4-4 \tan ^{2} \frac{x}{2}}{4+4 \tan ^{2} \frac{x}{2}}\right)+\frac{4 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} \\ \end{aligned}

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

\begin{aligned} &{\left[\begin{array}{l} \tan \frac{x}{2}=t \\ \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t \\ \sec ^{2} \frac{x}{2} d x=2 d t \end{array}\right]} \\ & \end{aligned}

I=\int \frac{2 d t}{4-4 t^{2}-4 t} \\

I=-\frac{1}{2} \int \frac{d t}{t^{2}-2 t-1}

\begin{aligned} I &=-\frac{1}{2} \int \frac{d t}{t^{2}+2 t-1} \\ \end{aligned}

I =-\frac{1}{2} \int \frac{d t}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}}

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

I=-\frac{1}{2 \sqrt{5}} \log \left|\frac{\tan \frac{x}{2}-\frac{1+\sqrt{5}}{2}}{\tan \frac{x}{2}-\frac{1-\sqrt{5}}{2}}\right|_{0}^{\frac{\pi}{2}}

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

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


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