Question

Evaluate the following integrals:

\[\int\left( x + 3 \right)\sqrt{3 - 4x - x^2} \text{  dx }\]

Answer 1

\[Let I = \int\left( x + 3 \right)\sqrt{3 - 4x - x^2} \text{  dx }\]
\[\text{ We  express  x + 3 }= A\left( \frac{d}{d x}\left( 3 - 4x - x^2 \right) \right) + B\]
\[x + 3 = A( - 4 - 2x) + B\]
\[\text{Equating the coefficients of x and constants, we get}\]
\[1 = - 2A \text{ and 3 } = - 4A + B\]
\[or A = - \frac{1}{2} \text{ and B} = 1 \]
\[ \therefore I = \int\left( - \frac{1}{2}\left( - 4 - 2x \right) + 1 \right)\sqrt{3 - 4x - x^2} \text{  dx }\]
\[ = - \frac{1}{2}\int\left( - 4 - 2x \right)\sqrt{3 - 4x - x^2} \text{  dx }+ \int\sqrt{3 - 4x - x^2} \text{  dx }\]
\[ = - \frac{1}{2} I_1 + I_2 . . . (1)\]
\[\text{ Now, }I_1 = \int\left( - 4 - 2x \right)\sqrt{3 - 4x - x^2} \text{  dx }\]
\[\text{ Let 3 }- 4x - x^2 = u\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( - 4 - 2x \right)dx = du\]
\[ \therefore I_1 = \int\sqrt{u}du\]
\[ = \frac{2}{3} u^\frac{3}{2} + c_1 \]
\[ = \frac{2}{3} \left( 3 - 4x - x^2 \right)^\frac{3}{2} + c_1 . . . (2)\]
\[\text{ And, I_2 }= \int\sqrt{3 - 4x - x^2} \text{  dx }\]
\[ = \int\sqrt{3 + 4 - 4 - 4x - x^2} \text{  dx }\]
\[ = \int\sqrt{\left( \sqrt{7} \right)^2 - \left( x + 2 \right)^2} \text{  dx }\]
\[ \text{ Let} \left( x + 2 \right) = u\]
\[ \text{On differentiating both sides, we get}\]
\[ dx = du\]
\[ \therefore I_2 = \int\sqrt{\left( \sqrt{7} \right)^2 - \left( u \right)^2} du\]
\[ = \frac{u}{2}\sqrt{\left( \sqrt{7} \right)^2 - \left( u \right)^2} + \frac{1}{2} \left( \sqrt{7} \right)^2 \sin^{- 1} \left( \frac{u}{\sqrt{7}} \right) + c_2 \]
\[ = \frac{x + 2}{2}\sqrt{7 - \left( x + 2 \right)^2} + \frac{7}{2} \sin^{- 1} \left( \frac{x + 2}{\sqrt{7}} \right) + c_2 \]
\[ = \frac{x + 2}{2}\sqrt{3 - 4x - x^2} + \frac{7}{2} \sin^{- 1} \left( \frac{x + 2}{\sqrt{7}} \right) + c_2 . . . (3)\]
\[\text{ From (1), (2) and (3), we get }\]
\[ \therefore I = - \frac{1}{2}\left( \frac{2}{3} \left( 3 - 4x - x^2 \right)^\frac{3}{2} + c_1 \right) + \left( \frac{x + 2}{2}\sqrt{3 - 4x - x^2} + \frac{7}{2} \sin^{- 1} \left( \frac{x + 2}{\sqrt{7}} \right) + c_2 \right)\]
\[ = - \frac{1}{3} \left( 3 - 4x - x^2 \right)^\frac{3}{2} + \frac{x + 2}{2}\sqrt{3 - 4x - x^2} + \frac{7}{2} \sin^{- 1} \left( \frac{x + 2}{\sqrt{7}} \right) + c\]
\[\text{ Hence,} \int\left( x + 3 \right)\sqrt{3 - 4x - x^2} \text{  dx }= - \frac{1}{3} \left( 3 - 4x - x^2 \right)^\frac{3}{2} + \frac{x + 2}{2}\sqrt{3 - 4x - x^2} + \frac{7}{2} \sin^{- 1} \left( \frac{x + 2}{\sqrt{7}} \right) + c\]