Question

`∫     cos ^4  2x   dx `

Answer 1

\[\int \cos^4 \text{2x dx}\]
\[ = \int \left( \cos^2 2x \right)^2 dx\]
\[ = \int \left( \frac{1 + \cos 4x}{2} \right)^2 dx \left[ \therefore \cos^2 x = \frac{1 + \cos 2x}{2} \right]\]
\[ = \frac{1}{4}\int \left( 1 + \cos 4x \right)^2 dx\]
\[ = \frac{1}{4}\int\left( 1 + \cos^2 4x + 2 \cos 4x \right)dx\]
\[ = \frac{1}{4}\int\left[ 1 + \left( \frac{1 + \cos 8x}{2} \right) + 2 \cos 4x \right]dx\]
\[ = \frac{1}{4}\int\left( \frac{3}{2} + \frac{\cos 8x}{2} + 2 \cos 4x \right)dx\]
\[ = \frac{1}{4}\left[ \frac{3x}{2} + \frac{\sin 8x}{16} + \frac{2 \sin 4x}{4} \right] + C\]

\[ = \frac{3x}{8} + \frac{\sin 8x}{64} + \frac{\sin 4x}{8} + C\]