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Q14  Integrate the functions \frac{1}{\sqrt { 8 + 3 x - x ^ 2 }}

Answers (1)

best_answer

We can write denominator as 
\\=8-(x^2-3x+\frac{9}{4}-\frac{9}{4})\\ =\frac{41}{4}-(x-\frac{3}{2})^2
 

therefore   
\Rightarrow \int \frac{1}{\sqrt{8+3x-x^2}}dx= \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^2}}
let x-3/2 = t \Rightarrow dx =dt

\therefore 
\\=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2-t^2}}dt\\ =\sin^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C\\ =\sin^{-1}(\frac{2x-3}{\sqrt{41}})+C

Posted by

manish

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