int frac{2 cos 3 x-3 sin 3 x}{cos 3 x+2 sin 3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x$.We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{den

Vedclass · Accepted Answer

$\frac{7}{15} \log |\cos 3 x+2 \sin 3 x|-\frac{4}{5} x+c$

Vedclass · Answer

(A) Let $I = \int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x$.We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator}))$.Let $2 \cos 3 x-3 \sin 3 x = A(\cos 3 x+2 \sin 3 x) + B(-3 \sin 3 x + 6 \cos 3 x)$.Equating coefficients of $\cos 3 x$ and $\sin 3 x$:$A + 6B = 2$ and $2A - 3B = -3$.Multiplying the second equation by $2$: $4A - 6B = -6$.Adding the two equations: $5A = -4 \Rightarrow A = -\frac{4}{5}$.Substituting $A$ into $A + 6B = 2$: $-\frac{4}{5} + 6B = 2 \Rightarrow 6B = \frac{14}{5} \Rightarrow B = \frac{7}{15}$.Thus, $I = \int \left( A + B \frac{-3 \sin 3 x + 6 \cos 3 x}{\cos 3 x+2 \sin 3 x} ight) d x$.$I = A \int 1 dx + B \int \frac{d(\cos 3 x+2 \sin 3 x)}{\cos 3 x+2 \sin 3 x}$.$I = -\frac{4}{5} x + \frac{7}{15} \log |\cos 3 x+2 \sin 3 x| + c$.

$\int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x=$

Similar Questions

$\int \frac{dx}{2 + \cos x} = $

$\int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx =$

$\int \frac{\sec x}{3(\sec x+\tan x)+2} d x=$

The value of $\frac{e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx}{\int_0^{\pi/4} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$ is

Evaluate the integral: $\int \sec^5 x \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module