Đặt : $t=a^2\cos^2 x+b^2\sin^2 x \Rightarrow dt=2(b^2-a^2)\sin x.\cos xdx$$TH_1 : a^2\neq b^2 \Rightarrow I=\frac{1}{2(b^2-a^2)}\int\limits_{a^2}^{b^2}\frac{dt}{{\sqrt{t}} }=\frac{1}{b^2-a^2}{\sqrt{t}}|^{b^2}_{a^2} =\frac{|b|-|a|}{b^2-a^2} =\frac{1}{|a|+|b|} (1)$$TH_2 :a^2=b^2$$\Rightarrow I=\int\limits_{0}^{\frac{\pi}{2} } \frac{\sin x.\cos x dx}{{\sqrt{a^2}} }=\frac{1}{|a|}\int\limits_{0}^{\frac{\pi}{2} }\sin xd(\sin x) = \frac{1}{2|a|}\sin^2 x|^{\frac{\pi}{2} }_0=\frac{1}{|2a|} =\frac{1}{|a|+|b|} (2) $Vậy $\forall a,b\neq 0. Từ (1) và (2) \Rightarrow I=\frac{1}{|a|+|b|}$
Đặt : $t=a^2\cos^2 x+b^2\sin^2 x \Rightarrow dt=2(b^2-a^2)\sin x.\cos xdx$$TH_1 : a^2\neq b^2 \Rightarrow I=\frac{1}{2(b^2-a^2)}\int\limits_{a^2}^{b^2}\frac{dt}{{\sqrt{t}} }=\frac{1}{b^2-a^2}{\sqrt{t}}|^{b^2}_{a^2} =\frac{|b|-|a|}{b^2-a^2} =\frac{1}{|a|+|b|} (1)$$TH_2 :a^2=b^2$$\Rightarrow I=\int\limits_{0}^{\frac{\pi}{2} } \frac{\sin x.\cos x dx}{{\sqrt{a^2}} }\frac{1}{|a|}\int\limits_{0}^{\frac{\pi}{2} }\sin xd(\sin x) = \frac{1}{2|a|}\sin^2 x|^{\frac{\pi}{2} }_0\frac{1}{|2a|} =\frac{1}{|a|+|b|} (2) $Vậy $\forall a,b\neq 0.Từ (1) và (2) \Rightarrow I=\frac{1}{|a|+|b|} (ycbt)$
Đặt : $t=a^2\cos^2 x+b^2\sin^2 x \Rightarrow dt=2(b^2-a^2)\sin x.\cos xdx$$TH_1 : a^2\neq b^2 \Rightarrow I=\frac{1}{2(b^2-a^2)}\int\limits_{a^2}^{b^2}\frac{dt}{{\sqrt{t}} }=\frac{1}{b^2-a^2}{\sqrt{t}}|^{b^2}_{a^2} =\frac{|b|-|a|}{b^2-a^2} =\frac{1}{|a|+|b|} (1)$$TH_2 :a^2=b^2$$\Rightarrow I=\int\limits_{0}^{\frac{\pi}{2} } \frac{\sin x.\cos x dx}{{\sqrt{a^2}} }
=\frac{1}{|a|}\int\limits_{0}^{\frac{\pi}{2} }\sin xd(\sin x) = \frac{1}{2|a|}\sin^2 x|^{\frac{\pi}{2} }_0
=\frac{1}{|2a|} =\frac{1}{|a|+|b|} (2) $Vậy $\forall a,b\neq 0.
Từ (1) và (2) \Rightarrow I=\frac{1}{|a|+|b|}$