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

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(A) To evaluate the integral $I = \int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx$,we express the numerator as a linear combination of the deno

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx$,we express the numerator as a linear combination of the denominator and its derivative: $3\sin x + 2\cos x = A(3\cos x + 2\sin x) + B \frac{d}{dx}(3\cos x + 2\sin x)$ $3\sin x + 2\cos x = A(3\cos x + 2\sin x) + B(-3\sin x + 2\cos x)$ Comparing the coefficients of $\sin x$ and $\cos x$ on both sides: For $\sin x$: $3 = 2A - 3B$ For $\cos x$: $2 = 3A + 2B$ Solving these simultaneous equations: Multiply the first by $2$ and the second by $3$: $6 = 4A - 6B$ $6 = 9A + 6B$ Adding them: $12 = 13A \implies A = \frac{12}{13}$ Substituting $A$ into the first equation: $3 = 2(\frac{12}{13}) - 3B \implies 3B = \frac{24}{13} - 3 = \frac{24 - 39}{13} = -\frac{15}{13} \implies B = -\frac{5}{13}$ Thus,the integral becomes: $I = \int \left( \frac{12}{13} \frac{3\cos x + 2\sin x}{3\cos x + 2\sin x} - \frac{5}{13} \frac{-3\sin x + 2\cos x}{3\cos x + 2\sin x} \right) dx$ $I = \frac{12}{13} \int 1 \, dx - \frac{5}{13} \int \frac{d}{dx}(3\cos x + 2\sin x) \cdot \frac{1}{3\cos x + 2\sin x} \, dx$ $I = \frac{12}{13}x - \frac{5}{13} \log |3\cos x + 2\sin x| + C$

$\int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx = $

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