Question

If $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C,(a>0, b>0)$ then $\int \frac{d x}{b+a \cos x}=$

• A. $\frac{2}{3} \log \left(\frac{3 \tan \frac{x}{2}+4-\sqrt{3}}{3 \tan \frac{x}{2}+4+\sqrt{3}}\right)+C$
• B. $\frac{2}{\sqrt{7}} \text{Tan}^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{7}}\right)+C$
• C. $\frac{2}{\sqrt{7}} \log \left(\frac{\sqrt{7}-\tan \frac{x}{2}}{\sqrt{7}+\tan \frac{x}{2}}\right)+C$
• D. $2 \text{Sinh}^{-1}\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+C$

Answer 1

Answer: (B)

We have
$\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} $
$\log |(b \sin x+5 \cos x)|+C$
Now, let $ a \cos x-2 \sin x=\lambda(b \sin x+5 \cos x)+\mu(b \cos x-5 \sin x) $
$\therefore a=5 \lambda+\mu b \text { and }-2=b \lambda-5 \mu$
But it is given, $ \lambda=\frac{7}{41} \text { and } \mu=\frac{22}{41} $
$\therefore a=\frac{35}{41}+\frac{22}{41} b \text { and }-2=\frac{7 b}{41}-\frac{110}{41} $
$a=3, b=4 $
$\int \frac{d x}{b+a \cos x}=\int \frac{d x}{4+3 \cos x}$
$=\int \frac{\left(1+\tan ^{2} \frac{x}{2}\right) d x}{4\left(1+\tan ^{2} \frac{x}{2}\right)+3\left(1-\tan ^{2} \frac{x}{2}\right)}$
$=\int \frac{\sec ^{2} x / 2}{7+\tan ^{2} x / 2} d x$
$\tan \frac{x}{2}=t \Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t$
$\frac{d x}{b+a \cos x}=2 \int \frac{d t}{7+t^{2}}$
$=\frac{2}{\sqrt{7}} \tan ^{-1} \frac{t}{\sqrt{7}}+C$
$=\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\tan x / 2}{\sqrt{7}}\right)+C$