Question
$\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}$
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Answer
Let I $=\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}.$ Then,
$\text{I}=\int\sin\text{x}\times\sqrt{2\sin^2\text{x}}\times\text{dx}$
$=\int\sin\text{x}\times\sqrt{2}\times\sin\text{x dx}$
$=\sqrt{2}\int\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\int2\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]+\text{C}$
$=\frac{\sqrt{2}\text{x}}{2}-\frac{\sqrt{2}}{4}\times\sin2\text{x}+\text{C}$
$=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
$\therefore\text{I}=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
$\text{I}=\int\sin\text{x}\times\sqrt{2\sin^2\text{x}}\times\text{dx}$
$=\int\sin\text{x}\times\sqrt{2}\times\sin\text{x dx}$
$=\sqrt{2}\int\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\int2\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]+\text{C}$
$=\frac{\sqrt{2}\text{x}}{2}-\frac{\sqrt{2}}{4}\times\sin2\text{x}+\text{C}$
$=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
$\therefore\text{I}=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
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