∫( 3 sin x + 2 cos x/3 cos x + 2 sin x) dx = ax + b ln | 2 sin x + 3 cos x | + C then

Question

∫( 3 sin x + 2 cos x/3 cos x + 2 sin x) dx = ax + b ln | 2 sin x + 3 cos x | + C then (A) a=-(12/13), b=(15/39) (B) a=-(7/13), b=(6/13) (C) a=(12/13), b=-(15/39) (D) a=-(7/13), b=-(6/13)

Tardigrade Admin · Accepted Answer

Differentiating both sides, we get (3 sin x + 2 cos x/3 cos x + 2 sin x) = a + (b(2 cos x - 3 sin x)/(2 sin x + 3 cos x)) = (sin x(2a - 3b)+ cos x ( 3a + 2b)/(3 cos x + 2 sin x)) Comparing like terms on both sides, we get3 = 2a-3b, 2 - 3a + 2b or a=(12/13), b=-(15/39)

Q. $\int \frac{3 s in x + 2 cos x}{3 cos x + 2 s in x} d x = a x + b l n ∣ 2 s in x + 3 cos x ∣ + C$ then

$a = - \frac{12}{13}, b = \frac{15}{39}$

$a = - \frac{7}{13}, b = \frac{6}{13}$

$a = \frac{12}{13}, b = - \frac{15}{39}$

$a = - \frac{7}{13}, b = - \frac{6}{13}$

Q. ∫3cosx+2sinx3sinx+2cosx​dx=ax+bln∣2sinx+3cosx∣+C then

a=−1312​,b=3915​

a=−137​,b=136​

a=1312​,b=−3915​

a=−137​,b=−136​

Differentiating both sides, we get 3cosx+2sinx3sinx+2cosx​=a+(2sinx+3cosx)b(2cosx−3sinx)​ =(3cosx+2sinx)sinx(2a−3b)+cosx(3a+2b)​ Comparing like terms on both sides, we get3=2a−3b,2−3a+2b or a=1312​,b=−3915​

Q. $\int \frac{3 s in x + 2 cos x}{3 cos x + 2 s in x} d x = a x + b l n ∣ 2 s in x + 3 cos x ∣ + C$ then

$a = - \frac{12}{13}, b = \frac{15}{39}$

$a = - \frac{7}{13}, b = \frac{6}{13}$

$a = \frac{12}{13}, b = - \frac{15}{39}$

$a = - \frac{7}{13}, b = - \frac{6}{13}$