Question

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

• 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
$\int \frac{a\cos x-2\sin x}{b\sin x+5\cos x}dx=\frac{7}{41}x+\frac{22}{41}$
$\log |(b\sin x+5\cos x)|+C$
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, $\lambda =\frac{7}{41}\text{ and }\mu =\frac{22}{41}$
$\therefore a=\frac{35}{41}+\frac{22}{41}b\text{ and }-2=\frac{7b}{41}-\frac{110}{41}$
$a=3,b=4$
$\int \frac{dx}{b+a\cos x}=\int \frac{dx}{4+3\cos x}$
$=\int \frac{\left(1+{\tan }^2\frac{x}{2}\right)dx}{4\left(1+{\tan }^2\frac{x}{2}\right)+3\left(1-{\tan }^2\frac{x}{2}\right)}$
$=\int \frac{{\sec }^{2}x/2}{7+{\tan }^{2}x/2}dx$
$\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$
$\frac{dx}{b+a\cos x}=2\int \frac{dt}{7+t^2}$
$=\frac{2}{\sqrt{7}}{\tan }^{-1}\frac{t}{\sqrt{7}}+C$
$=\frac{2}{\sqrt{7}}{\tan }^{-1}\left(\frac{\tan x/2}{\sqrt{7}}\right)+C$