Search any question & find its solution
Question:
Answered & Verified by Expert
If $\int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x=A \log$
$|5 \cos x+4 \sin x|+B x+c$, then $A$ and $B$ are
Options:
$|5 \cos x+4 \sin x|+B x+c$, then $A$ and $B$ are
Solution:
1852 Upvotes
Verified Answer
The correct answer is:
$A=\frac{22}{41}$ and $B=\frac{7}{41}$
We have,
$\int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x$
$=A \log (5 \cos x+4 \sin x)+B x+c$
On differentiating both sides, we get
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x}=\frac{A(-5 \sin x+4 \cos x)}{4 \sin x+5 \cos x}+B$
$\Rightarrow \quad \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x}$
$=\frac{-5 A \sin x+4 A \cos x+4 B \sin x+5 B \cos x}{4 \sin x+5 \cos x}$
$\Rightarrow 3 \cos x-2 \sin x=(4 B-5 A) \sin x+(5 B+4 A) \cos x$
Equating the coefficient of $\cos x$ and $\sin x$, we get
$5 B+4 A=3$ and $5 A-4 B=2$
Solving, we get
$A=\frac{22}{41}$ and $B=\frac{7}{41}$
$\int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x$
$=A \log (5 \cos x+4 \sin x)+B x+c$
On differentiating both sides, we get
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x}=\frac{A(-5 \sin x+4 \cos x)}{4 \sin x+5 \cos x}+B$
$\Rightarrow \quad \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x}$
$=\frac{-5 A \sin x+4 A \cos x+4 B \sin x+5 B \cos x}{4 \sin x+5 \cos x}$
$\Rightarrow 3 \cos x-2 \sin x=(4 B-5 A) \sin x+(5 B+4 A) \cos x$
Equating the coefficient of $\cos x$ and $\sin x$, we get
$5 B+4 A=3$ and $5 A-4 B=2$
Solving, we get
$A=\frac{22}{41}$ and $B=\frac{7}{41}$
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.