giai giup minh pai nay nha.thanks!

Question

$\int\limits_{0}^{\frac{\pi}{2}}\frac{(3sinx+4cosx)dx}{3sin^{2}x+4cos^{2}x}$

score 1 · Answer 1

=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) =

$-\int\limits_{0}^{1}(\frac{1}{v-2}-\frac{1}{v+2}).dv =-ln\left|v-2 {} \right|\begin{cases}1 \\ 0 \end{cases} + ln\left|v+2 {} \right|\begin{cases}1 \\ 0 \end{cases}=ln3$