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  • Provide Solution For RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 7 Maths Textbook Solution.

Provide Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 7 Maths Textbook Solution.

Answers (1)

Answer : \frac{1}{6} \tan ^{-1}\left[\frac{5 \tan \frac{x}{2}+2}{6}\right]+c

Hint: To solve this equation we have to convert cosx and sinx in terms of tanx.

Given:  \int \frac{1}{13+3 \cos x+4 \sin x} d x

Solution : I=\int \frac{1}{13+3 \cos x+4 \sin x} d x

We will use the formula : \left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]

\begin{aligned} &I=\int \frac{1}{13+3\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+4\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{13+13 \tan ^{2} \frac{x}{2}+3-3 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \end{aligned}

I=\int \frac{\sec ^{2} \frac{x}{2}}{10 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}+16} d x

\begin{aligned} &=\int \frac{\sec ^{2} \frac{x}{2}}{2\left(5 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+8\right)} d x \\ &\text { put } t=\tan \frac{x}{2} \\ &\frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \\ &I=\int \frac{d t}{\left(5 t^{2}+4 t+8\right)} \end{aligned}

\begin{aligned} &=\int \frac{d t}{5\left(t^{2}+\frac{4}{5}+\frac{8}{5}\right)} \\ &=\int \frac{d t}{5\left(t^{2}+2 \frac{2}{5} t+\left(\frac{2}{5}\right)^{2}-\left(\frac{2}{5}\right)^{2}+\frac{8}{5}\right)} \\ &=\frac{1}{5} \int \frac{d t}{\left(t+\frac{2}{5}\right)^{2}+\frac{36}{25}} \end{aligned}

=\frac{1}{5} \int \frac{d u}{\left(t+\frac{2}{5}\right)^{2}+\left(\frac{6}{5}\right)^{2}}

Use the formula : \left[\int\left(\frac{d x}{x^{2}+a^{2}}\right)=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]

\begin{aligned} &I=\frac{1}{5} \times \frac{5}{6} \tan ^{-1}\left(\frac{\left(t+\frac{2}{5}\right)}{\frac{6}{5}}\right)+c \\ &I=\frac{1}{6} \tan ^{-1}\left(\frac{\frac{5 t+2}{5}}{\frac{6}{5}}\right)+c \\ &I=\frac{1}{6} \tan ^{-1}\left(\frac{5 \tan \frac{x}{2}+2}{6}\right)+c \end{aligned}

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