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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 58

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 58

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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)

 

Posted by

infoexpert27

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