Question

If $\int \frac{d x}{5+4 \cos x}=A \tan ^{-1}\left(B \tan \frac{x}{2}\right)+C$, then

(a) $A=1$

(b) $A=\frac{2}{3}$

(c) $B=\frac{5}{3}$

(d) $B=\frac{2}{3}$

Answer 1

$\text { Let } I =\int \frac{1}{5+4 \cos x} d x \\$

$ =\int \frac{1+\tan ^2 x / 2}{5\left(1+\tan ^2 x / 2\right)+4\left(1-\tan ^2 x / 2\right)} d x$

Put $\tan \frac{x}{2}=t \Rightarrow \sec ^2 \frac{x}{2} \cdot \frac{1}{2} d x=d t$

$\therefore \quad I=\int \frac{2 d t}{t^2+9}=\frac{2}{3} \tan ^{-1}\left(\frac{t}{3}\right)+C \\$

$=\frac{2}{3} \tan ^{-1}\left(\frac{\tan (x / 2)}{3}\right)+C \\$

$\therefore \quad A=\frac{2}{3} \text { and } B=\frac{1}{3} \\$

So, The correct option of this question will be (B).