Question

integral of sin^4x/cos^2x.dx ​

Answer 1

Answer:

\begin{eqnarray}

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

&=& \frac{1}{4} \int (1-2\cos^2x+\cos^4x)\cos^2x dx \\

&=& \frac{1}{4} \int \cos^6x-2\cos^4x+\cos^2xdx \\

&=& \frac{1}{4} \left( \frac{1}{7}\cos^7x-\frac{2}{5}\cos^5x+\frac{1}{3}\cos^3x \right) + C

\end{eqnarray}