_ ^ 3 x + 2 x 3 x + 2 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}\;dx = } $

✓
$\frac{{12}}{{13}}x - \frac{5}{{13}}\log (3\cos x + 2\sin x)$
B
$\frac{{12}}{{13}}x + \frac{5}{{13}}\log (3\cos x + 2\sin x)$
C
$\frac{{13}}{{12}}x + \frac{5}{{13}}\log (3\cos x + 2\sin x)$
D
None of these

✓

Answer

Correct option: A.

$\frac{{12}}{{13}}x - \frac{5}{{13}}\log (3\cos x + 2\sin x)$

a
(a) Write ${N^r} = l({D^r}) + m$ (differential coefficient of ${D^r}).$
Let $3sinx+2cosx = l(3cosx+2sinx)+m(û3sinx + 2cosx)$
Comparing coefficients of $\sin x$ and $\cos x$on both sides $3 = 2l - 3m$ and $2 = 3l + 2m$
Solving, we get $l = \frac{{12}}{{13}},$ $m = - \frac{5}{{13}},$
$\therefore \,\,I = l\,\int_{}^{} {dx} + m\int_{}^{} {\frac{{ - 3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}\,dx} $
=$lx+mlog(3cosx+2sinx)$ $ = \frac{{12}}{{13}}x - \frac{5}{{13}}$ $log(3cosx+2sinx).$

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