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Q12  Integrate the functions \frac{1}{\sqrt{ 7-6x - x ^ 2 }}

Answers (1)

best_answer

the denominator can be also written as,
7-6x-x^2=16-(x^2+6x+9)
                           =4^2-(x+3)^2

therefore

\int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{4^2-(x+3)^2}}dx
                                               Let x+3 = t 
                                                then dx =dt 

                                            \Rightarrow \int \frac{1}{\sqrt{4^2-(x+3)^2}}dx=\int \frac{1}{\sqrt{4^2-t^2}}dt......................................using formula \int \frac{1}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})
                                                                                                   \\= sin^{-1}(\frac{t}{4})+C\\ =\sin^{-1}(\frac{x+3}{4})+C

Posted by

manish

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