Question

Find the integral of sin^2xcos^4xdx

Answer 1

\[\int sin^2xcos^4x dx\]
Now make them linear I mean
\[\int \frac{4(sin^2x.cos^2x) cos^2x}{4}\]

Answer 2

\[\frac{1}{4}\int sin^22x \times \frac{(cos2x - 1 )}{2}\]

Answer 3

\[\frac{1}{8}\int sin^22x cos2x - sin^22x dx \]

Answer 4

cos^2 x=[cos2x+1]/2
not [cos2x-1]/2

Answer 5

Then first integral can be solved by t = sin2x substitution

Answer 6

\[\frac{1}{8} ( \frac{1}{2} \times \frac{1}{3} t^3 ) + \frac{1}{8}\int sin^22xdx\]

Answer 7

Now you can convert sin^2x into linear too

Answer 8

\[sin^22x = \frac{1 - cos4x}{2}\]

Answer 9

\[\frac{1}{8 * 2 *3}sin^32x + \frac{1}{8*2}x - \frac{1}{8*2*4}sin4x +C\]