The value of int frac{d x}{5+4 sin x} is equa | vedclass

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(B) Let $I = \int \frac{dx}{5+4 \sin x}$.Using the substitution $	an(\frac{x}{2}) = t$,we have $dx = \frac{2 dt}{1+t^2}$ and $\sin x = \frac{2t}{1+t^

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$\frac{2}{3} 	an ^{-1}\left(\frac{5 	an \frac{x}{2}+4}{3}ight)+c$,(where $c$ is a constant of integration)

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(B) Let $I = \int \frac{dx}{5+4 \sin x}$.Using the substitution $	an(\frac{x}{2}) = t$,we have $dx = \frac{2 dt}{1+t^2}$ and $\sin x = \frac{2t}{1+t^2}$.Substituting these into the integral:$I = \int \frac{\frac{2 dt}{1+t^2}}{5 + 4(\frac{2t}{1+t^2})} = \int \frac{2 dt}{5(1+t^2) + 8t} = 2 \int \frac{dt}{5t^2 + 8t + 5}$.Factor out $5$ from the denominator:$I = \frac{2}{5} \int \frac{dt}{t^2 + \frac{8}{5}t + 1}$.Complete the square in the denominator:$t^2 + \frac{8}{5}t + 1 = (t + \frac{4}{5})^2 + 1 - \frac{16}{25} = (t + \frac{4}{5})^2 + \frac{9}{25} = (t + \frac{4}{5})^2 + (\frac{3}{5})^2$.Now,use the standard integral formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + c$:$I = \frac{2}{5} \cdot \frac{1}{3/5} 	an^{-1}(\frac{t + 4/5}{3/5}) + c = \frac{2}{3} 	an^{-1}(\frac{5t + 4}{3}) + c$.Substituting $t = 	an(\frac{x}{2})$ back:$I = \frac{2}{3} 	an^{-1}\left(\frac{5 	an(\frac{x}{2}) + 4}{3}ight) + c$.

The value of $\int \frac{d x}{5+4 \sin x}$ is equal to

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