Question

$\int \frac{1}{5+4\cos x}dx=$

• A. $\frac{2}{3}{\tan }^{-1}(\frac{\tan \frac{x}{2}}{3})+c$
• B. $\frac{1}{3}{\tan }^{-1}(\frac{\tan \frac{x}{2}}{3})+c$
• C. ${\tan }^{-1}(\frac{\tan \frac{x}{2}}{3})+c$
• D. $\frac{2}{3}{\tan }^{-1}\left(\begin{array}{c}\tan -x\\ 2\end{array}\right)+c$

Answer 1

The correct option is A $\frac{2}{3}{\tan }^{-1}(\frac{\tan \frac{x}{2}}{3})+c$

$\int \frac{1}{5+4cosx}dx=\int \frac{dx}{5+4(1-2si{n}^2{}^x{/}_2)}$
$=\int \frac{dx}{5+4(2co{s}^2{}^x{/}_2-1)}=\int \frac{d{(}^x{/}_2)}{1+8co{s}^2{}^x{/}_2}$
$\int \frac{se{c}^2{}^x{/}_{2}dx}{se{c}^2{}^x{/}_2+8}=\int \frac{se{c}^2{}^x{/}_{2}dx}{ta{n}^2{}^x{/}_2+9}$$ta{n}^x{/}_2=t$
$se{c}^2{}^x{/}_{2}dx=2dt$
$\int \frac{2dt}{t^2+9}=\frac{2}{3}Ta{n}^{-1}\left(\frac{t}{3}\right)+c$
$=\frac{2}{3}Ta{n}^{-1}\left(\frac{tan{}^x{/}_2}{3}\right)+c$