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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
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Answer

Consider the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$Let us express $\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+3\text{x}-18)+\mu$
$\Rightarrow\text{x}-3=\lambda(2\text{x}+3)+\mu$
$\Rightarrow\text{x}-3=2\lambda\text{x}+3\lambda+\mu$
Comparing the co-efficients, we have,
$2\lambda=1\text{ and }3\lambda+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }3\times\frac{1}{2}+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu-3-\frac{3}{2}$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu=-\frac{9}{2}$
Then
$\text{x}-3=\lambda(2\text{x}+3)+\mu$
Now the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=\int\Big(\frac{1}{2}(2\text{x}+3)-\frac{9}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\text{I}=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}\\-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\Rightarrow\text{I}=\text{I}_1+\text{I}_2$
where, $\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
and $\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Let us consider the integral, $I_1$:
$\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Substituting, $\text{x}^2+3\text{x}-18=\text{t}$
$\Rightarrow(2\text{x}+3)\text{dx = dt}$
Thus,
$\text{I}_1=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=\frac{1}{2}\times\frac{2}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}+\text{C}$
Now consider the integral
$\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\text{x}^2+2\times\frac{3}{2}\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\frac{9}{4}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{4}+18\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9+72}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{81}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\text{dx}$
We know that,
$\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}-\frac{1}{2}\text{a}^2\log\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}$
$\therefore\ \text{I}_2=-\frac{9}{2}\begin{Bmatrix}\frac{1}{2}\Big(\text{x}+\frac{3}{2}\Big)\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\\-\frac{1}{2}\Big(\frac{9}{2}\Big)^2\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{4}\begin{Bmatrix}\Big(\frac{2\text{x}+3}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}-\Big(\frac{729}{4}\Big)\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}+\frac{729}{16}\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
Thus,
$\text{I}=-\frac{1}{3}(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\\+\frac{729}{16}\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$

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