Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{d x}{4 \sin x+3 \cos x}=$
Options:
Solution:
1489 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{5} \log \left|\sec \left(x-\tan ^{-1} \frac{4}{3}\right)+\tan \left(x-\tan ^{-1} \frac{4}{3}\right)\right|+c$
$I=\int \frac{d x}{4 \sin x+3 \cos x}$
Let $3=r \cos \theta$ and $4=r \sin \theta$
$\Rightarrow \quad r=5$ and $\theta=\tan ^{-1} \frac{4}{3}$
So, $I=\int \frac{d x}{r \cos (x-\theta)}=\frac{1}{5} \int \sec (x-\theta) d \theta$
$\begin{aligned} & =\frac{1}{5} \log |\sec (x-\theta)+\tan (x-\theta)|+c \\ & =\frac{1}{5} \log \left|\sec \left(x-\tan ^{-1} \frac{4}{3}\right)+\tan \left(x-\tan ^{-1} \frac{4}{3}\right)\right|+c\end{aligned}$
Hence, option (c) is correct.
Let $3=r \cos \theta$ and $4=r \sin \theta$
$\Rightarrow \quad r=5$ and $\theta=\tan ^{-1} \frac{4}{3}$
So, $I=\int \frac{d x}{r \cos (x-\theta)}=\frac{1}{5} \int \sec (x-\theta) d \theta$
$\begin{aligned} & =\frac{1}{5} \log |\sec (x-\theta)+\tan (x-\theta)|+c \\ & =\frac{1}{5} \log \left|\sec \left(x-\tan ^{-1} \frac{4}{3}\right)+\tan \left(x-\tan ^{-1} \frac{4}{3}\right)\right|+c\end{aligned}$
Hence, option (c) is correct.
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.