If int_0^{pi / 2} frac{d x}{5+4 sin x}=A tan | vedclass

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(B) Let $I=\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}$.Substitute $	an \frac{x}{2}=t$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dt$,so $dx = \frac{2 dt}

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$1$

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(B) Let $I=\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}$.Substitute $	an \frac{x}{2}=t$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dt$,so $dx = \frac{2 dt}{1+t^2}$ and $\sin x = \frac{2t}{1+t^2}$.When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$.$I = \int_0^1 \frac{1}{5+4(\frac{2t}{1+t^2})} \cdot \frac{2 dt}{1+t^2} = 2 \int_0^1 \frac{dt}{5+5t^2+8t} = \frac{2}{5} \int_0^1 \frac{dt}{t^2+\frac{8}{5}t+1}$.Completing the square: $t^2+\frac{8}{5}t+1 = (t+\frac{4}{5})^2 + (1-\frac{16}{25}) = (t+\frac{4}{5})^2 + (\frac{3}{5})^2$.$I = \frac{2}{5} \int_0^1 \frac{dt}{(t+\frac{4}{5})^2 + (\frac{3}{5})^2} = \frac{2}{5} \cdot \frac{1}{3/5} [	an^{-1}(\frac{t+4/5}{3/5})]_0^1 = \frac{2}{3} [	an^{-1}(\frac{5t+4}{3})]_0^1$.$I = \frac{2}{3} [	an^{-1}(3) - 	an^{-1}(\frac{4}{3})] = \frac{2}{3} 	an^{-1}(\frac{3-4/3}{1+3(4/3)}) = \frac{2}{3} 	an^{-1}(\frac{5/3}{5}) = \frac{2}{3} 	an^{-1}(\frac{1}{3})$.Comparing with $A 	an^{-1} B$,we get $A=\frac{2}{3}$ and $B=\frac{1}{3}$.Therefore,$A+B = \frac{2}{3} + \frac{1}{3} = 1$.

If $\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B$,then $A+B=$

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