#### Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17Question 7 Maths Textbook Solution

Answer:- $\sin^{-1}\left ( \frac{x+3}{5} \right )+c$

Hint: - To solve this problem, use special integration formula

Given:- $\int \frac{1}{\sqrt{16-6x-x^{2}}}dx$

Solution:- $Let\: I=\int \frac{1}{\sqrt{16-6x-x^{2}}}dx\Rightarrow \int \frac{1}{\sqrt{-\left \{ x^{2}+6x-16 \right \}}}dx$

\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{x^{2}+2 x .3+(3)^{2}-(3)^{2}-16\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{(x+3)^{2}-9-16\right\}}} d x=\int \frac{1}{\sqrt{-\left\{(x+3)^{2}-25\right\}}} d x \\ &\Rightarrow I=\int \frac{1}{\sqrt{25-(x+3)^{2}}} d x=\int \frac{1}{\sqrt{(5)^{2}-(x+3)^{2}}} d x \end{aligned}

\Gamma \begin{aligned} &\text { put } x+3=t \Rightarrow d x=d t \text { then }\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]\\&I=\int \frac{1}{\sqrt{(5)^{2}-t^{2}}} d t=\sin ^{-1}\left(\frac{t}{5}\right)+c\\ &=\sin ^{-1}\left(\frac{x+3}{4}\right)+c \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad[\because t=x+3] \end{aligned}