int frac{dx}{4+3 cot x} = | vedclass

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(B) Let $I = \int \frac{dx}{4+3 \cot x} = \int \frac{\sin x dx}{4 \sin x + 3 \cos x}$.We express the numerator as $A(4 \cos x - 3 \sin x) + B(4 \sin x

Vedclass · Accepted Answer

$-\frac{3}{25} \log |4 \sin x+3 \cos x|+\frac{4}{25} x+c$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{4+3 \cot x} = \int \frac{\sin x dx}{4 \sin x + 3 \cos x}$.We express the numerator as $A(4 \cos x - 3 \sin x) + B(4 \sin x + 3 \cos x)$, where $4 \cos x - 3 \sin x$ is the derivative of the denominator $4 \sin x + 3 \cos x$.Equating coefficients of $\sin x$ and $\cos x$:$4B - 3A = 1$ and $3B + 4A = 0$.From $3B + 4A = 0$, we get $A = -\frac{3B}{4}$.Substituting into the first equation: $4B - 3(-\frac{3B}{4}) = 1 \Rightarrow 4B + \frac{9B}{4} = 1 \Rightarrow \frac{25B}{4} = 1 \Rightarrow B = \frac{4}{25}$.Then $A = -\frac{3}{25}$.Thus, $I = \int \frac{-\frac{3}{25}(4 \cos x - 3 \sin x) + \frac{4}{25}(4 \sin x + 3 \cos x)}{4 \sin x + 3 \cos x} dx$.$I = -\frac{3}{25} \int \frac{4 \cos x - 3 \sin x}{4 \sin x + 3 \cos x} dx + \frac{4}{25} \int dx$.$I = -\frac{3}{25} \ln |4 \sin x + 3 \cos x| + \frac{4}{25} x + C$.

$\int \frac{dx}{4+3 \cot x} = $

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