Question

Find the integrals of the functions.
$\int si{n}^{4}xdx.$

Answer 1

\(\int sin^4 x dx=\int (sin^2 x)^2 dx\\
=\int \left ( \frac{1-cos 2x}{2} \right )^2 dx =\frac{1}{4}\int (1-cos 2x)^2 dx \right]\left ( \therefore sin^2 x=\frac{1-cos 2x}{2} \right ) \\

=\frac{1}{4}\int 1+cos^2 2x -2 cos 2x dx]\\
=\frac{1}{4}[\int 1 dx +\int cos^2 2x dx -2 \int cos 2x dx]\\
=\frac{1}{4}\left [ \int 1 dx +\int \frac{1+cos 4x}{2}dx -2 \int cos 2x dx \right ] \left ( \therefore cos^2 x=\frac{1+cos 2x}{2} \right )\\
=\frac{1}{4}\left [ \int 1 dx +\frac{1}{2}(\int 1 dx +\int cos 4x dx)-2 \int cos 2x dx \right ]\\
=\frac{1}{4}\left [ x+\frac{1}{2}\left \{ x+\frac{sin 4x}{4} \right \}-\frac{2 sin 2x}{2} \right ]+C\\
=\frac{1}{4}x+\frac{x}{8}+\frac{sin 4x}{32}-\frac{sin 2x}{4}+C =\left ( \frac{2x+x}{8} \right )+\frac{sin 4x}{32}-\frac{sin 2x}{4}+C\\
=\left [ \frac{3x}{8}+\frac{sin 4x}{32}-\frac{sin 2x}{4} \right ]+C\)