If int frac{dx}{2 cos x + 3 sin x + 4} = frac | vedclass

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(C) Let $I = \int \frac{dx}{2 \cos x + 3 \sin x + 4}$.Using the Weierstrass substitution,let $t = 	an\left(\frac{x}{2}ight)$,then $dx = \frac{2 dt}

Vedclass · Accepted Answer

$\frac{5 \pi}{12}$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{2 \cos x + 3 \sin x + 4}$.Using the Weierstrass substitution,let $t = 	an\left(\frac{x}{2}ight)$,then $dx = \frac{2 dt}{1+t^2}$,$\cos x = \frac{1-t^2}{1+t^2}$,and $\sin x = \frac{2t}{1+t^2}$.Substituting these into the integral:$I = \int \frac{\frac{2 dt}{1+t^2}}{2\left(\frac{1-t^2}{1+t^2}ight) + 3\left(\frac{2t}{1+t^2}ight) + 4} = \int \frac{2 dt}{2 - 2t^2 + 6t + 4 + 4t^2} = \int \frac{2 dt}{2t^2 + 6t + 6} = \int \frac{dt}{t^2 + 3t + 3}$.Completing the square in the denominator: $t^2 + 3t + 3 = \left(t + \frac{3}{2}ight)^2 + \frac{3}{4}$.Thus,$I = \int \frac{dt}{\left(t + \frac{3}{2}ight)^2 + \left(\frac{\sqrt{3}}{2}ight)^2} = \frac{1}{\frac{\sqrt{3}}{2}} 	an^{-1}\left(\frac{t + 3/2}{\sqrt{3}/2}ight) + c = \frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{2t+3}{\sqrt{3}}ight) + c$.Comparing this with the given form $\frac{2}{\sqrt{3}} f(x) + c$,we get $f(x) = 	an^{-1}\left(\frac{2 	an(x/2) + 3}{\sqrt{3}}ight)$.Now,evaluate $f\left(\frac{2 \pi}{3}ight) = 	an^{-1}\left(\frac{2 	an(\pi/3) + 3}{\sqrt{3}}ight) = 	an^{-1}\left(\frac{2\sqrt{3} + 3}{\sqrt{3}}ight) = 	an^{-1}(2 + \sqrt{3})$.Since $	an(75^\circ) = 	an\left(\frac{5 \pi}{12}ight) = 2 + \sqrt{3}$,we have $f\left(\frac{2 \pi}{3}ight) = \frac{5 \pi}{12}$.

If $\int \frac{dx}{2 \cos x + 3 \sin x + 4} = \frac{2}{\sqrt{3}} f(x) + c$,then $f\left(\frac{2 \pi}{3}\right) =$

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