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  • Provide Solution For RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 10 Maths Textbook Solution.

Provide Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 10 Maths Textbook Solution.

Answers (1)

Answer : \frac{2}{3} \tan ^{-1}\left[3 \tan \frac{x}{2}\right]+c

Hint: To solve this equation we have to change cosx into tanx form

Given: \int \frac{1}{5-4 \cos x} d x

Solution :\int \frac{1}{5-4 \cos x}dx

\operatorname{Cos} x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

\begin{aligned} &=\int \frac{1}{5-4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} \mathrm{d} x \\ &= \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \end{aligned}

\begin{aligned} &=\int \frac{\sec ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{1+\left(3 \tan \frac{x}{2}\right)^{2}} d x \end{aligned}

\begin{aligned} &\text { Let } 3 \tan \frac{x}{2}=t \\ &=3 \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}

\begin{aligned} &=\frac{2}{3} \int \frac{d t}{1+t^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\langle\int\frac{1}{a^{2}+t^{2}}dt=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)\right\rangle \\ &=\frac{2}{3} \tan ^{-1}(t)+C \\ &=\frac{2}{3} \tan ^{-1}\left(3 \tan \frac{x}{2}\right)+C \end{aligned}

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