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Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17 Question 8 Maths Textbook Solution

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Answer:- \sin^{-1}\left ( \frac{x+3}{4} \right )+c

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

Given:- \int \frac{1}{\sqrt{7-6x-x^{2}}}dx


Let I=\int \frac{1}{\sqrt{7-6x-x^{2}}}dx\Rightarrow \int \frac{1}{\sqrt{-\left \{ x^{2}+6x -7\right \}}}dx 

        \begin{aligned} &=\int \frac{1}{\sqrt{-\left\{x^{2}+2 x \cdot 3+(3)^{2}-(3)^{2}-7\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{(x+3)^{2}-9-7\right\}}} d x \Rightarrow \int \frac{1}{\sqrt{-\left\{(x+3)^{2}-16\right\}}} d x \\ &=\int \frac{1}{\sqrt{16-(x+3)^{2}}} d x \Rightarrow \int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x \\ &\text { put } \mathrm{x}+3=\mathrm{t} \Rightarrow \mathrm{dx}=\mathrm{dt} \text { then } \end{aligned}

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


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