Question
Integrate the functions in Exercises:
$\sqrt{\sin2\text{x}}\cos2\text{x}$
$\sqrt{\sin2\text{x}}\cos2\text{x}$
✓
Answer
$\text{Let I}=\int\sqrt{\sin2\text{x }}\cos 2\text{x dx}=\frac{1}{2}\int\sqrt{\sin2\text{x }}(2\cos\text{2x dx}) \ \ \ \ \ \ \ ...\text{(i)} $
Putting $\sin2\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ \ \cos2 \text{x}\frac{\text{d}}{\text{dx}}(2\text{x})=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow\ \ \ 2\cos 2\text{x dx = dt} $
$\therefore\ \ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{2}\int\text{t}^{\frac{1}{2}}\text{ dt}=\frac{1}{2}\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c}$
$=\frac{1}{2}\dot\ \frac{\text{t}^{^3/_2}}{^3/_2}+\text{c}=\frac{1}{3}(\sin 2\text{x})^{\frac{3}{2}}+\text{c} $
Putting $\sin2\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ \ \cos2 \text{x}\frac{\text{d}}{\text{dx}}(2\text{x})=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow\ \ \ 2\cos 2\text{x dx = dt} $
$\therefore\ \ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{2}\int\text{t}^{\frac{1}{2}}\text{ dt}=\frac{1}{2}\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c}$
$=\frac{1}{2}\dot\ \frac{\text{t}^{^3/_2}}{^3/_2}+\text{c}=\frac{1}{3}(\sin 2\text{x})^{\frac{3}{2}}+\text{c} $
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