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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:$\int\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}\text{ dx}$
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Answer

$8 + 3x - x^2$ can be written as $8-\Big(\text{x}^2-3\text{x}+\frac{9}{4}-\frac{9}{4}\Big).$Therefore,
$8-\Big(\text{x}^2-3\text{x}+\frac{9}{4}-\frac{9}{4}\Big)$
$=\frac{41}{4}-\Big(\text{x}-\frac{3}{2}\Big)^2$
$=\int\frac{1}{8+3\text{x}-\text{x}^2}\text{ dx}$
$=\int\frac{1}{\sqrt{\frac{41}{4}-\big(\text{x}-\frac{3}{2}\big)^2}}\text{ dx}$
Let $\text{x}-\frac{3}{2}=\text{t}$
$\therefore\ \text{dx}=\text{dt}$
$=\int\frac{1}{\sqrt{\frac{41}{4}-\big(\text{x}-\frac{3}{2}\big)^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big(\frac{\sqrt{41}}{2}\Big)^2-\text{t}^2}}\text{ dt}$
$=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}$
$=\sin^{-1}\bigg(\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}$
$=\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{41}}\Big)+\text{C}$

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