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Need Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 13 Maths Textbook Solution.

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Answer : \frac{1}{2} \log \left|\sec \left(x-\frac{\pi}{3}\right)+\tan \left(x-\frac{\pi}{3}\right)\right|+c

Hint: To solve this question we have to use formula of \cos \left(x-\frac{\pi}{3}\right)

Given: \int \frac{1}{\sqrt{3} \sin x+\cos x} d x

Solution : Multiplying and dividing the denominator by 2

\begin{aligned} &=\int \frac{1}{2\left(\frac{\sqrt{3}}{2} \sin x+\left(\frac{1}{2}\right) \cos x\right)} d x \\ &=\frac{1}{2} \int \frac{1}{\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x} d x \\ &=\frac{1}{2} \int \frac{1}{\cos \frac{\pi}{3} \cos x+\sin \frac{\pi}{3} \sin x} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \cos (\mathrm{A}-\mathrm{B})=\cos A \cos \mathrm{B}+\sin A \sin \mathrm{s} \\ &=\frac{1}{2} \int \frac{1}{\cos \left(x-\frac{\pi}{3}\right)} d x \end{aligned}

\begin{aligned} &=\frac{1}{2} \int \sec \left(x-\frac{\pi}{3}\right) d x \\ &t=x-\frac{\pi}{3} \\ &\frac{d t}{d x}=1 \\ &d t=d x \end{aligned}

\begin{aligned} &=\frac{1}{2} \int \sec t d t \\ &=\frac{1}{2} \log |\sec t+\tan t|+c \\ &=\frac{1}{2} \log \left|\sec \left(x-\frac{\pi}{3}\right)+\tan \left(x-\frac{\pi}{3}\right)\right|+c \end{aligned}

Note : Final answer is not matching with the book.

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