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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 50 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 50 Maths Textbook Solution.

 

Answers (1)

Answer:

I=\sin ^{-1}\left(\frac{x+1}{2}\right)+c

Given:

\int \frac{1}{\sqrt{3-2 x-x^{2}}} d x

Hint

 

Solution: 

\int \frac{1}{\sqrt{3-2 x-x^{2}}} d x

  I=\int \frac{1}{\sqrt{3+1-(x+1)^{2}}} d x

I=\int \frac{1}{\sqrt{4-(x+1)^{2}}} d x

I=\int \frac{d x}{\sqrt{(2)^{2}-(x+1)^{2}}}

t=x+1

d t=d x

=\int \frac{d t}{\sqrt{2^{2}-t^{2}}} \quad\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+c\right.

=\sin ^{-1}\left(\frac{t}{2}\right)+c

I=\sin ^{-1}\left(\frac{x+1}{2}\right)+c

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