Question

lf $\int \frac{2\sin x+3\cos x}{3\sin x+4\cos x}dx=A$ Iog $|3\sin x+4\cos x|+$ Bx $+c$, then $A=\dots \dots \dots ,B=\dots \dots \dots .$,

• A. $-\frac{1}{25},\frac{18}{25}$
• B. $-\frac{1}{5},-\frac{1}{5}$
• C. $\frac{1}{25},\frac{18}{25}$
• D. $\frac{1}{25},\frac{3}{25}$

Answer 1

Answer: (C)

$\frac{1}{25},\frac{18}{25}$

We can write, 2 sin x + 3 cos x in a different form as,
$2sinx+3cosx=\lambda (3sinx+4cosx)+\mu (3cosx-4sinx)$
Comparing the coefficient of sin x and cos x,
$\lambda =\frac{18}{25}and\mu =\frac{1}{25}$
Thus, the integral can be written as,
$\int \frac{18}{25}\times \frac{3sinx+4cosx}{3sinx+4cosx}dx+\frac{1}{25}\times \frac{3cosx-4sinx}{3sinx+4cosx}dx$
= $\int \frac{18}{25}dx+\int \frac{1}{25}\times \frac{3cosx-4sinx}{3sinx+4cosx}dx$
Let $3sinx+4cosx=t$
Then,
$3cosx-4sinxdx=dt$
Hence, $\int \frac{18}{25}dx+\int \frac{1}{25}\times \frac{dt}{t}$
$=\frac{18}{25}x+\frac{1}{25}logt+c$
Putting the value of $t$,
$\frac{18}{25}x+\frac{1}{25}log(3sinx+4cosx)+c$
Comparing it with the RHS,
$A=\frac{1}{25}andB=\frac{18}{25}$