Find the integrals of the functions in Exercises: ^4x - Vidyadip

Question

Find the integrals of the functions in Exercises:
$\sin^4\text{x}$

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Answer

$\sin^4\text{x}=\sin^2\text{x}\sin^2\text{x}$
$=\bigg(\frac{1-\cos2\text{x}}{2}\bigg)\bigg(\frac{1-\cos2\text{x}}{2}\bigg)$
$=\frac{1}{4}(1-\cos2\text{x})^2$
$=\frac{1}{4}\Big[1+\cos^22\text{x}-2\cos2\text{x}\Big]$
$=\frac{1}{4}\Bigg[1+\bigg(\frac{1+\cos4\text{x}}{2}\bigg)-2\cos2\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[1+\frac{1}{2}+\frac{1}{2}\cos4\text{x}-2\cos2\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[\frac{3}{2}+\frac{1}{2}\cos4\text{x}-2\cos2\text{x}\Bigg]$
$\therefore\int\sin^4\text{x}\text{ dx}=\frac{1}{4}\int\bigg[\frac{3}{2}+\frac{1}{2}\cos4\text{x}-2\cos2\text{x}\bigg]\text{ dx}$
$=\frac{1}{4}\bigg[\frac{3}{2}\text{x}+\frac{1}{2}\bigg(\frac{\sin4\text{x}}{4}\bigg)-\frac{2\sin2\text{x}}{2}\bigg]+\text{C}$
$=\frac{1}{8}\bigg[3\text{x}+\frac{\sin4\text{x}}{4}-2\sin2\text{x}\bigg]+\text{C}$
$=\frac{3\text{x}}{8}-\frac{1}{4}\sin2\text{x}+\frac{1}{32}\sin4\text{x}+\text{C}$

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