INTEGRALS — Maths STD 12 Science — Question

Let $(2\text{x}+5)=\text{A}\frac{\text{d}}{\text{dx}}(10-4\text{x}-3\text{x}^2)+\text{B}$

$\Rightarrow(2\text{x}+5)=\text{A}(-4-6\text{x})+\text{B}$

$\Rightarrow(2\text{x}+5)=-6\text{A}\text{x}+(\text{B}-4\text{A})$

$\Rightarrow2=-6\text{A}\text{ and }(\text{B}-4\text{A})=5$

$\Rightarrow\text{A}=-\frac{1}{3}\text{ and }\text{B}=\frac{11}{3}$

$\Rightarrow(2\text{x}+5)=-\frac{1}{3}(-4-6\text{x})+\frac{11}{3}$

$\Rightarrow\text{I}=-\frac{1}{3}(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}\\+\frac{11}{3}\int\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$

Let $\text{I}=-\frac{1}{3}\text{I}_1+\frac{11}{3}\text{I}_2\ \dots(1)$

Now,

$\text{I}_1=\int(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$

Let $(10-4\text{x}-3\text{x}^2)=\text{t},\text{ or, }(-4-6\text{x})\text{dx = dt}$

$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$

$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$

$\Rightarrow\text{I}_1=\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\text{c}_1$

$\text{I}_2=\int\sqrt{(10-4\text{x}-3\text{x}^2)}\text{dx}$

$=\int\sqrt{3\Big(\frac{10}{3}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$

$=\sqrt3\int\sqrt{\Big(\frac{26}{9}-\frac{4}{9}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$

$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\frac{4}{9}+\frac{4}{3}\text{x}-\text{x}^2\Big)\Big]}\text{dx}$

$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\text{x}+\frac{2}{3}\Big)^2\Big]}\text{dx}$

$=\sqrt3\sin\Bigg(\frac{\text{x}+\frac{2}{3}}{\frac{\sqrt{26}}{3}}\Bigg)+\text{c}_2$

$=\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{c}_2$

Using (1), we get

$\text{I}=-\frac{1}{3}\times\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}\\+\frac{11}{3}\times\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$

$\therefore\ \text{I}=-\frac{2}{9}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\frac{11\sqrt3}{3}\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$