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Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int(2\text{x}-5)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
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Answer

Let $\text{I}=\int(2\text{x}-5)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
Also, $2\text{x}-5=\lambda\frac{\text{d}}{\text{dx}}(2+3\text{x}-\text{x}^2)+\mu$
$\Rightarrow2\text{x}-5=\lambda(-2\text{x}+3)+\mu$
$\Rightarrow2\text{x}-5=(-2\lambda)\text{x}=3\lambda+\mu$
Equating co-efficients of like terms
$-2\lambda=2$
$\Rightarrow\lambda=-1$
And
$3\lambda+\mu=-5$
$\Rightarrow3(-1)+\mu=-5$
$\Rightarrow\mu=-5+3$
$\Rightarrow\mu=-2$
$\therefore\ 2\text{x}-5=-1(-2\text{x}+3)-2$
Hence, $\text{I}=\int[-(-2\text{x}+3)-2]\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
$=-\int(-2\text{x}+3)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}-2\int\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
$=-\text{I}_1-2\text{I}_2\ \dots(1)$
$\text{I}_1=\int(-2\text{x}+3)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
Let $2+3\text{x}-\text{x}^2=\text{t}$
$\Rightarrow(-2\text{x}+3)\text{dx}=\text{dt}$
$\therefore\ \text{I}_1=\int\text{t}^{\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}$
$=\frac{2}{3}\big(2+3\text{x}-\text{x}^2\big)^{\frac{3}{2}}\ \dots(2)$
And $\text{I}_2=\int\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
$\text{I}_2=\int\sqrt{2-(\text{x}^2-3\text{x})}\text{dx}$
$=\int\sqrt{2-\Big[\text{x}^2-3\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2\Big]}\text{dx}$
$=\int\sqrt{2+\frac{9}{4}-\Big(\text{x}-\frac{3}{2}\Big)^2}\text{dx}$
$=\int\sqrt{\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}-\frac{3}{2}\Big)^2}\text{dx}$
$=\frac{\text{x}-\frac{3}{2}}{2}\sqrt{\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}-\frac{3}{2}\Big)^2}+\frac{\Big(\frac{\sqrt{17}}{2}\Big)^2}{2}\sin^{-1}\Bigg(\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{17}}{2}}\Bigg)$
$=\frac{2\text{x}-3}{4}\sqrt{2+3\text{x}-\text{x}^2}+\frac{17}{8}\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{17}}\Big)\ \dots(3)$
From eq. (1), (2) and (3) we have
$\text{I}=-\frac{2}{3}\big(2+3\text{x}-\text{x}^2\big)^{\frac{3}{2}}-\frac{(2\text{x}-3)}{2}\sqrt{2+3\text{x}-\text{x}^2}\\-\frac{17}{4}\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{17}}\Big)+\text{C}$

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