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=y=$\int\limits_{0}^{\frac{\Pi}{2}}\frac{3\sin x.dx}{3+ \cos ^{2}x} + \int\limits_{0}^{\frac{\Pi}{2}}\frac{4\cos x.dx}{4-\sin ^{2}x}$
=$\int\limits_{0}^{1}\frac{3.dt}{3 + t^{2}} (1) với t= cosx; dt=-sinx.dx; \begin{cases}0 \rightarrow1 \\ \frac{\Pi}{2}\rightarrow0 \end{cases}$
+$ \int\limits_{0}^{1}\frac{4.du}{4-u^2} (2) với u = sinx; du = cosx.dx; \begin{cases}0\rightarrow 0\\ \frac{\Pi}{2}\rightarrow1 \end{cases}$
(1) =$\int\limits_{0}^{\frac{\Pi}{6}} \frac{1}{1+\tan ^2 v}.\frac{\sqrt{3}}{cos^2v}.dv; với t=\sqrt{3}tanv; dt = \frac{\sqrt{3}}{cos^2v}.dv; \begin{cases}0\rightarrow0 \\ 1\rightarrow \frac{\Pi}{6} \end{cases}$
=$\sqrt{3}. \frac{\Pi}{6} $
(2) = $-\int\limits_{0}^{1}(\frac{1}{v-2}-\frac{1}{v+2}).dv =-ln\left|v-2 {} \right|\begin{cases}1 \\ 0 \end{cases} + ln\left|v+2 {} \right|\begin{cases}1 \\ 0 \end{cases}=ln3$
$\Rightarrow y= \sqrt{3}.\frac{\Pi}{6} + ln3$
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