Question

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then

विकल्प

•  A =\[\frac{2}{3}\], B =\[\frac{5}{3}\]

•  A =\[\frac{1}{3}\], B = \[\frac{2}{3}\]

•  A =\[- \frac{2}{3}\], B =\[\frac{5}{3}\]

• A =\[\frac{1}{3}\], B =\[- \frac{5}{3}\]

Answer 1

A =\[\frac{2}{3}\] , B =\[\frac{5}{3}\]

\[\int\frac{1}{5 + 4 \sin x}dx =\text{ A  }\tan^{- 1} \left( \text{ B} \tan \frac{x}{2} + \frac{4}{3} \right) + C . . . . (1)\]
\[\text{Considering the LHS of eq} \text{ (1)}\]
\[\text{ Putting  sin x }= \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \Rightarrow \int\frac{1}{5 + \frac{8 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ \Rightarrow \int\frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{5 \left( 1 + \tan^2 \frac{x}{2} \right) + 8 \tan \frac{x}{2}}\text{ dx }\]
\[ \Rightarrow \int\frac{\sec^2 \frac{x}{2}}{5 \tan^2 \frac{x}{2} + 8 \tan \frac{x}{2} + 5}\text{  dx }. . . (2) \]
\[\text{ Let tan }\frac{x}{2} = t\]
\[ \Rightarrow \sec^2 \frac{x}{2} \times \frac{1}{2} \text{ dx }= dt\]
\[ \Rightarrow \sec^2 \left( \frac{x}{2} \right)\text{  dx }= 2dt\]
\[ \therefore \text{ Putting  tan} \frac{x}{2} = \text{ t  and }\sec^2 \left( \frac{x}{2} \right) dx = \text{ 2dt we get, }\]
\[\int\frac{2dt}{5 t^2 + 8t + 5}\]
\[ \Rightarrow \frac{2}{5}\int\frac{dt}{t^2 + \frac{8}{5}t + 1}\]
\[ \Rightarrow \frac{2}{5}\int\frac{dt}{t^2 + \frac{8}{5}t + \left( \frac{4}{5} \right)^2 - \left( \frac{4}{5} \right)^2 + 1}\]
\[ \Rightarrow \frac{2}{5}\int\frac{dt}{\left( t + \frac{4}{5} \right)^2 + 1 - \frac{16}{25}}\]
\[ \Rightarrow \frac{2}{5}\int\frac{dt}{\left( t + \frac{4}{5} \right)^2 + \left( \frac{3}{5} \right)^2}\]
\[ \Rightarrow \frac{2}{5} \times \frac{5}{3} \tan^{- 1} \left( \frac{t + \frac{4}{5}}{\frac{3}{5}} \right) + C\]
\[ \Rightarrow \frac{2}{3} \text{ tan}^{- 1} \left( \frac{5t + 4}{3} \right) + C\]
\[ \Rightarrow \frac{2}{3} \text{ tan}^{- 1} \left( \frac{5}{3} \tan \frac{x}{2} + \frac{4}{3} \right) + C \left( \because t = \text{ tan} \frac{x}{2} \right) . . . (3)\]

\[\text{ Comparing eq (3) with the RHS of eq (1) we get ,} \]
\[ \therefore A = \frac{2}{3}, B = \frac{5}{3}\]