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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 16 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 16 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+C

Given:

\int\left(\frac{1}{7+5 \cos x}\right) d x

Hint:

You must know about the derivation of tan x and using \int \frac{1}{1+x^{2}} d x

Explanation:

Let \mathrm{I}=\int \frac{1}{7+5 \cos x} d x

\text { Put } \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

\begin{aligned} \therefore I &=\int \frac{1}{7+5\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{7\left(1+\tan ^{2} \frac{x}{2}\right)+5-5 \tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{12+2 \tan ^{2} \frac{x}{2}} \\ &=\frac{1}{2} \int \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}+6} d x \\ &=\frac{1}{2} \int \frac{2 d t}{t^{2}+(\sqrt{6})^{2}} \end{aligned}    \text { [ Put } \left.\tan \frac{x}{2}=t \Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t\right]

        \begin{aligned} &=\int \frac{d t}{t^{2}+(\sqrt{6})^{2}} \\ &=\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{t}{\sqrt{6}}\right)+C \ldots \ldots \ldots \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \\ &=\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{6}}\right)+C \end{aligned}                                                

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