Evaluate Int Sin4 X Cos4X Dx

Sol:

Given \( \int sin^{4}x * cos^{4}x * dx \) \( I = \int sin^{4}x * cos^{4}x * dx \)

As we know,

\( sin^{2}x = \frac{1 – cos 2x}{2}, cos^{2}x = \frac{1 + cos 2x}{2} \) \( \int \left (\frac{1 – cos 2x}{2}\right )^{2} * \left (\frac{1 + cos 2x}{2} \right )^{2} dx \)

=\( \frac{1}{16} \int (1 – cos^{2} 2x)^{2} dx \)

=\( \frac{1}{16} \int 1 + \frac{(1 + cos4x)^{^{2}}}{4} – 2 \left ( \frac{1 + cos4x}{2} \right) dx \)

=\( \frac{1}{16} \int 1 + \frac{1 +cos^{2}+ 2 cos 4x}{2} 1 – cos4x dx \)

= \( \frac{1}{16} \int \frac{1 + \frac{1 + cos8x}{2}+ 2 cos 4x – 2 cos 4x}{2} dx \)

=\( \frac{1}{32} \int 1 + \frac{1 + cos8x}{2} dx \)

=\( \frac{1}{64} \int 3 + cos8x dx \)

=\( \frac{1}{64}\left [ 3x + \frac{sin 8x}{8} \right ] + C \)

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