int frac{d x}{4 sin x+3 cos x}= | vedclass

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(C) $I = \int \frac{dx}{4 \sin x + 3 \cos x}$Let $3 = r \cos 	heta$ and $4 = r \sin 	heta$.Squaring and adding,we get $r^2 = 3^2 + 4^2 = 25$,so $r =

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$\frac{1}{5} \log \left|\sec \left(x-	an ^{-1} \frac{4}{3}ight)+	an \left(x-	an ^{-1} \frac{4}{3}ight)ight|+c$

Vedclass · Answer

(C) $I = \int \frac{dx}{4 \sin x + 3 \cos x}$Let $3 = r \cos 	heta$ and $4 = r \sin 	heta$.Squaring and adding,we get $r^2 = 3^2 + 4^2 = 25$,so $r = 5$.Dividing,we get $	an 	heta = \frac{4}{3}$,so $	heta = 	an^{-1} \frac{4}{3}$.The integral becomes $I = \int \frac{dx}{r(\sin x \cos 	heta + \cos x \sin 	heta)} = \int \frac{dx}{r \sin(x + 	heta)}$.Alternatively,using $4 \sin x + 3 \cos x = 5(\frac{4}{5} \sin x + \frac{3}{5} \cos x) = 5 \cos(x - 	heta)$ where $\cos 	heta = \frac{3}{5}$ and $\sin 	heta = \frac{4}{5}$,which implies $	heta = 	an^{-1} \frac{4}{3}$.Thus,$I = \int \frac{dx}{5 \cos(x - 	heta)} = \frac{1}{5} \int \sec(x - 	heta) dx$.Using the formula $\int \sec u du = \log |\sec u + 	an u| + c$,we get:$I = \frac{1}{5} \log |\sec(x - 	heta) + 	an(x - 	heta)| + c$.Substituting $	heta = 	an^{-1} \frac{4}{3}$,we get $I = \frac{1}{5} \log |\sec(x - 	an^{-1} \frac{4}{3}) + 	an(x - 	an^{-1} \frac{4}{3})| + c$.

$\int \frac{d x}{4 \sin x+3 \cos x}=$

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