int frac{sin x+8 cos x}{4 sin x+6 cos x} d x | vedclass

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

Question

(A) Let $I = \int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$.We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator

Vedclass · Accepted Answer

$x+\frac{1}{2} \log |4 \sin x+6 \cos x|+c$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$.We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator}))$:$\sin x+8 \cos x = A(4 \sin x+6 \cos x) + B(4 \cos x-6 \sin x)$.Equating the coefficients of $\sin x$ and $\cos x$:For $\sin x$: $4A - 6B = 1$.For $\cos x$: $6A + 4B = 8$.Solving these equations,we multiply the first by $2$ and the second by $3$:$8A - 12B = 2$ and $18A + 12B = 24$.Adding them gives $26A = 26$,so $A = 1$.Substituting $A=1$ into $4A - 6B = 1$ gives $4 - 6B = 1$,so $6B = 3$,$B = \frac{1}{2}$.Thus,$I = \int \frac{(4 \sin x+6 \cos x) + \frac{1}{2}(4 \cos x-6 \sin x)}{4 \sin x+6 \cos x} d x$.$I = \int 1 d x + \frac{1}{2} \int \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x} d x$.$I = x + \frac{1}{2} \log |4 \sin x+6 \cos x| + c$.

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to

Similar Questions

The value of $\int {\frac{{{e^x} + 9\cos x - 2\sin x + 7}}{{{e^x} + 7\sin x + 11\cos x + 14}}\,dx} $ is (where $C$ is the constant of integration).

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

Evaluate: $\int \frac{2 \cos x+1}{(2+\cos x)^2} d x - \frac{\sin x}{2+\cos x}$

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

$\int \frac{1}{3 \cos x - 4 \sin x + 5} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module