Question

$\int \frac{dx}{5+4\cos x}=k{\tan }^{-1}\left(m\tan \frac{x}{2}\right)+C$ then?

• A. $k=2/3$
• B. $m=4/3$
• C. $k=1/3$
• D. $m=2/3$

Answer 1

The correct option is A $k=2/3$

$\int \frac{dx}{5+4\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}\Rightarrow \int \frac{dx}{\frac{\left(5+5{\tan }^2\frac{x}{2}+4-4{\tan }^2\frac{x}{2}\right)}{\left(1+{\tan }^2\frac{x}{2}\right)}}=k{\tan }^{-1}\left(m\tan \frac{x}{2}\right)+C\Rightarrow \int \frac{\left(1+{\tan }^2\frac{x}{2}\right)}{\left({\tan }^2\frac{x}{2}+9\right)}dx\Rightarrow \int \frac{{\sec }^2\frac{x}{2}dx}{\left({\tan }^2\frac{x}{2}+9\right)}\left[\because if{\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}\right]let\tan \frac{x}{2}=t\left({\sec }^2\frac{x}{2}\right)\frac{1}{2}dx=dtdx=\frac{2dt}{{\sec }^2\frac{x}{2}}\Rightarrow 2\int \frac{{\sec }^2\frac{x}{2}\times \frac{dt}{{\sec }^2\frac{x}{2}}}{\left(t^2+3^2\right)}\Rightarrow \int \frac{dt}{\left(t^2+3^2\right)}\left[\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c\right]\Rightarrow 2\times \frac{1}{3}ta{b}^{-1}\left(\frac{t}{3}\right)+c=k{\tan }^{-1}\left(\frac{1}{3}\cdot \tan \frac{x}{2}\right)+c\Rightarrow \frac{2}{3}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)+cbycomparingk=\frac{2}{3},andm=\frac{1}{3}$