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  • Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17 Question 3 Maths Textbook Solution

Please Solve RD Sharma Class 12 Chapter 18.17 Indefinite Integrals Exercise Very Short Answer Question 3 Maths Textbook Solution

Answers (1)

Answer:- \frac{1}{\sqrt{2}}\sin^{-1}\left ( \frac{x+1}{\sqrt{\frac{}{2}}} \right )+c                                                                                                                                                

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

Given:- \int \frac{1}{\sqrt{5-4x-2x^{2}}}dx

Solution:- Let \int \frac{1}{\sqrt{5-4x-2x^{2}}}dx\Rightarrow \int \frac{1}{\sqrt{-\left \{ \left ( 2x^{2} +4x-5\right ) \right \}}}dx  

\begin{gathered} =\int \frac{1}{\sqrt{-2\left\{x^{2}+2 x-\frac{5}{2}\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{x^{2}+2 x \cdot 1+(1)^{2}-(1)^{2}-\frac{5}{2}\right\}}} d x \\ \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-1-\frac{5}{2}\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{(x+1)^{2}-\left(1+\frac{5}{2}\right)\right\}}} d x \end{gathered}

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

Put  x+1=t\Rightarrow dx=dt then

I=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-t^{2}}} d t \Rightarrow \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{t}{\frac{\sqrt{7}}{2}}\right)+c \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]\\ = \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+1}{\frac{\sqrt{7}}{2}}\right)+c \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad{[\because t=x+1]} \quad \begin{aligned} \end{aligned}                          

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