int frac{d x}{12 cos x+5 sin x}= | vedclass

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(B) To evaluate the integral $I = \int \frac{dx}{12 \cos x + 5 \sin x}$,we express the denominator in the form $R \cos(x - \alpha)$.Let $12 \cos x + 5

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$\frac{1}{13} \log \left|	an \left(\frac{x}{2} - \frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12}ight)ight|+c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{dx}{12 \cos x + 5 \sin x}$,we express the denominator in the form $R \cos(x - \alpha)$.Let $12 \cos x + 5 \sin x = R \cos(x - \alpha) = R \cos x \cos \alpha + R \sin x \sin \alpha$.Comparing coefficients,$R \cos \alpha = 12$ and $R \sin \alpha = 5$.Squaring and adding,$R^2 = 12^2 + 5^2 = 144 + 25 = 169$,so $R = 13$.Also,$	an \alpha = \frac{5}{12}$,which implies $\alpha = \operatorname{Tan}^{-1} \frac{5}{12}$.Thus,$I = \int \frac{dx}{13 \cos(x - \alpha)} = \frac{1}{13} \int \sec(x - \alpha) dx$.The integral of $\sec 	heta$ is $\log |	an(\frac{	heta}{2} + \frac{\pi}{4})|$.Therefore,$I = \frac{1}{13} \log \left| 	an \left( \frac{x - \alpha}{2} + \frac{\pi}{4} ight) ight| + c$.Substituting $\alpha = \operatorname{Tan}^{-1} \frac{5}{12}$,we get $I = \frac{1}{13} \log \left| 	an \left( \frac{x}{2} - \frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12} + \frac{\pi}{4} ight) ight| + c$.

$\int \frac{d x}{12 \cos x+5 \sin x}=$

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