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  • Please solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 2 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 2 maths textbook solution.

Answers (1)

Answer : \frac{2}{3} \tan ^{-1}\left[\frac{5 \tan \frac{x}{2}-4}{3}\right]+c

Hint: To solve this equation we have to use \sin ^{2} x+\cos ^{2} x=1 and divide numerator and denominator by \cos ^{2} \frac{x}{2}

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

Solution :

\begin{aligned} &I=\int \frac{1}{5\left(\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}\right)-4\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} d x \; \; \; \; \; \; \; \; \; \; \quad\left[\sin ^{2} x+\cos ^{2} x=1, \sin 2 x=2 \sin x \cos x\right] \\ &=\int \frac{d x}{5 \sin ^{2} \frac{x}{2}+5 \cos ^{2} \frac{x}{2}-8 \sin \frac{x}{2} \cos \frac{x}{2}} \end{aligned}

Divide numerator and denominator by \cos ^{2} \frac{x}{2}

I=\int \frac{\sec ^{2} \frac{x}{2}}{5 \tan ^{2} \frac{x}{2}+5-8 \tan \frac{x}{2}} d x

Let

\begin{aligned} &\tan \frac{x}{2}=t \\ &\frac{\sec ^{2} \frac{x}{2}}{2} d x=d t \\ &I=2 \int \frac{1}{5 t^{2}+5-8 t} d t \\ &I=2 \times \frac{1}{5} \int \frac{1}{t^{2}-\frac{8 t}{5}+1} d t \end{aligned}

\begin{aligned} &I=\frac{2}{5} \int \frac{d t}{t^{2}-\frac{8 t}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} \\ &I=\frac{2}{5} \int \frac{d t}{\left(t-\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} \end{aligned}

\begin{aligned} &I=\frac{2}{5} \times \frac{5}{3} \tan ^{-1}\left|\frac{t-\frac{4}{5}}{\frac{3}{5}}\right|+c \\ &I=\frac{2}{3} \tan ^{-1}\left[\frac{5 t-4}{3}\right]+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right] \\ &I=\frac{2}{3} \tan ^{-1}\left[\frac{5 \tan \frac{x}{2}-4}{3}\right]+c \end{aligned}

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