Search any question & find its solution
Question:
Answered & Verified by Expert
If $\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+c$, then $a$ (Where $\mathrm{c}$ is constant of integration)
Options:
Solution:
2855 Upvotes
Verified Answer
The correct answer is:
2
Let $I=\int \frac{5 \tan x}{\tan x-2} d x$
$$
I=\int \frac{5 \sin x}{\sin x-2 \cos x} d x
$$
Here $\frac{d}{d x}(\sin x-2 \cos x)=\cos x+2 \sin x$
$$
\begin{aligned}
\therefore \mathrm{I} & =\int \frac{(2 \sin x+2 \sin x+\sin x)+(2 \cos x-2 \cos x)}{\sin x-2 \cos x} d x \\
& =\int \frac{(2 \sin x+\cos x)+(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \\
& =\int \frac{2(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \\
& =\int d x+2 \int \frac{2 \sin x+\cos x}{\sin x-2 \cos x} d x \\
& =x+2 \log |\sin x-2 \cos x|+c
\end{aligned}
$$
From given data, $a=2$
$$
I=\int \frac{5 \sin x}{\sin x-2 \cos x} d x
$$
Here $\frac{d}{d x}(\sin x-2 \cos x)=\cos x+2 \sin x$
$$
\begin{aligned}
\therefore \mathrm{I} & =\int \frac{(2 \sin x+2 \sin x+\sin x)+(2 \cos x-2 \cos x)}{\sin x-2 \cos x} d x \\
& =\int \frac{(2 \sin x+\cos x)+(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \\
& =\int \frac{2(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \\
& =\int d x+2 \int \frac{2 \sin x+\cos x}{\sin x-2 \cos x} d x \\
& =x+2 \log |\sin x-2 \cos x|+c
\end{aligned}
$$
From given data, $a=2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.