Question

Integrate the following.

1. ∫sin x sin 2x sin 3 xdx

2. ∫sec²x cos²2x dx

Answer 1

1. We have sinxsin2xsin3x

= \(\frac{1}{2}\) (2sinxsin3x) sin2x

= \(\frac{1}{2}\) (cos2x – cos4x) sin2x

= \(\frac{1}{4}\) (2sin2xcos2x – 2cos4xsi n2x)

= \(\frac{1}{4}\) [sin4x – (sin6x – sin2x)]

= \(\frac{1}{4}\)(sin4x + sin2x – sin6x)

∫sin x sin 2x sin 3 xdx

= \(\frac{1}{4}\) ∫(sin 4x + sin 2x – sin 6x)dx

= –\(\frac{1}{16}\) cos4x – \(\frac{1}{8}\) cos2x + \(\frac{1}{24}\) cos6x + c.

2. sec²x cos²2x = \((\frac{2cos^2x -1}{cos^2x})^2\)

\((\frac{2cos^2x}{cos^2x} - \frac{1}{cosx})^2\) = (2cosx – secx)²

= 4cos²x + sec²x – 4

= 2(1 + cos2x) + sec²x – 4

= 2cos2x + sec²x – 2

∫sec² x cos² 2x dx = ∫(2 cos 2x + sec² x – 2)dx

= sin 2x + tan x – 2x + c.