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  • Please solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 4 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 4 maths textbook solution.

Answers (1)

Answer : \frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{x}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{x}{2}}\right|+c

Hint: To solve this statement we have to convert cos into tan form

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

Solution :

\int \frac{1}{4 \cos x-1} d x

Use the formula : \left[\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right]

\begin{aligned} &=\int \frac{1}{4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)-1} d x \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{4-4 \tan ^{2} \frac{x}{2}-1-\tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{3-5 \tan ^{2} \frac{x}{2}} d x \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} \tan \frac{x}{2}=t \\ \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t \end{array}\right]} \\ &=\int \frac{2 d t}{3-5 t^{2}} \\ &=\frac{1}{\sqrt{3}} \int \frac{2 \sqrt{3}}{(\sqrt{3}-\sqrt{5} t)(\sqrt{3}+\sqrt{5} t)} d t \\ &=\frac{1}{\sqrt{3}} \int \frac{(\sqrt{3}-\sqrt{5} t)+(\sqrt{3}+\sqrt{5} t)}{(\sqrt{3}-\sqrt{5} t)(\sqrt{3}+\sqrt{5} t)} d t \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{3}} \int \frac{d t}{\sqrt{3}+\sqrt{5} t}+\frac{1}{\sqrt{3}} \int \frac{d t}{\sqrt{3}-\sqrt{5} t}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{a t+b}=\log \left|\frac{a t+b}{a}\right|+c\right] \\ &=\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3}+\sqrt{5} t}{\sqrt{5}}\right|+\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3}-\sqrt{5} t}{-\sqrt{5}}\right|+c \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{3}+\sqrt{5} t}{\sqrt{3}-\sqrt{5} t}\right|+c \\ &=\frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{x}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{x}{2}}\right|+C \end{aligned}

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