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Question: Answered & Verified by Expert
$\int \frac{d x}{4 \sin x+3 \cos x}=$
MathematicsIndefinite IntegrationTS EAMCETTS EAMCET 2019 (06 May Shift 1)
Options:
  • A $\frac{1}{5} \log \left|\sec \left(x-\tan ^{-1} \frac{4}{3}\right)\right|+c$
  • B $\frac{1}{5} \log \left|\tan \left(\frac{\pi}{4}-x+\tan ^{-1} \frac{4}{3}\right)\right|+c$
  • C $\frac{1}{5} \log \left|\sec \left(x-\tan ^{-1} \frac{4}{3}\right)+\tan \left(x-\tan ^{-1} \frac{4}{3}\right)\right|+c$
  • D $\frac{1}{5} \log \left|\operatorname{cosec}\left(x-\tan ^{-1} \frac{4}{3}\right)+\cot \left(x-\tan ^{-1} \frac{4}{3}\right)\right|+c$
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.

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