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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
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Answer

Let $\text{I}=\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$We express $\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(3-4\text{x}-\text{x}^2)+\text{B}$
$\text{x}+3=\text{A}(-4-2\text{x})+\text{B}$
Equating the co-efficient of x and constants, we get
$1=-2\text{A}$ and $3=-4\text{A + B}$
or $\text{A}=-\frac{1}{2}$ and $\text{B}=1$
$\therefore\ \text{I}=\int\Big(-\frac{1}{2}(-4-2\text{x})+1\Big)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=-\frac{1}{2}\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}+\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=-\frac{1}{2}\text{I}_1+\text{I}_2\ \dots(1)$
Now, $\text{I}_1=\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
Let $3-4\text{x}-\text{x}^2=\text{u}$
On differentiating both sides, we get
$(-4-2\text{x})\text{dx = du}$
$\therefore\ \text{I}_1=\int\sqrt{\text{u}}\text{du}$
$=\frac{2}{3}\text{u}^{\frac{3}{2}}+\text{c}_1$
$=\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\ \dots(2)$
And, $\text{I}_2=\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{3+4-4-4\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{(\sqrt7)^2-(\text{x}+2)^2}\text{dx}$
Let $(\text{x}+2)=\text{u}$
On differentiating both sides, we get
$\text{dx = du}$
$\therefore\ \text{I}_2=\int\sqrt{(\sqrt7)^2-(\text{u})^2}\text{du}$
$=\frac{\text{u}}{2}\sqrt{(\sqrt7)^2-(\text{u})^2}+\frac{1}{2}(\sqrt7)^2\sin^{-1}\Big(\frac{\text{u}}{\sqrt7}\Big)+\text{c}_2$
$=\frac{\text{x}+2}{2}\sqrt{7-(\text{x}+2)^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2$
$=\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\ \dots(3)$
From (1), (2) and (3), we get
$\therefore\ \text{I}=-\frac{1}{2}\Big(\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\Big)\\+\Big(\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\Big)$
$=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}\\+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$
Hence,
$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}\\+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$

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