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  • Need Solution For RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 15 Maths Textbook Solution.

Need Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 15 Maths Textbook Solution.

Answers (1)

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

Hint: To solve this question we have to use partial equation

Given:  \int \frac{1}{5+7 \cos x+\sin x} d x

Solution :

I=\int \frac{1}{5+7\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x                                                 \sin 2 \theta=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \cos 2 \theta=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

\begin{aligned} &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+7\left(1-\tan ^{2} \frac{x}{2}\right)+2 \tan \frac{x}{2}} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+7-7 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}} d x \end{aligned}

\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{12-2 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}} d x\\ &=\frac{1}{2} \int \frac{\sec ^{2} \frac{x}{2}}{6-\tan ^{2} \frac{x}{2}+\tan \frac{x}{2}} d x \end{aligned}

\begin{aligned} &\text { Put } \tan \frac{x}{2}=t \\ &\ \sec ^{2} \frac{x}{2}\cdot\frac{1}{2} d x=d t \\ &I=\int \frac{d t}{6-t^{2}+t} \\ &=\int \frac{-d t}{t^{2}-t-6} \\ &=\int \frac{-d t}{t^{2}-3 t+2 t-6} \end{aligned}

\begin{aligned} &=\int \frac{-d t}{t(t-3)+2(t-3)} \\ &=-\frac{1}{5} \int \frac{(3+2) d t}{(t-3)(t+2)} \\ &=-\frac{1}{5} \int \frac{[(t+2)-(t-3)] d t}{(t-3)(t+2)} \\ &=-\frac{1}{5} \int \frac{d t}{(t-3)}+\frac{1}{5} \int \frac{d t}{(t+2)} \end{aligned}

\begin{aligned} &=\frac{1}{5}[\log |t+2|-\log |t-3|]+C \\ &=\frac{1}{5} \log \left|\frac{t+2}{t-3}\right|+C \\ &=\frac{1}{5} \log \left|\frac{\tan \frac{x}{2}+2}{\tan \frac{x}{2}-3}\right|+C \end{aligned}

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