Question

Integrate:
$\int {\cos }^2(4x+3)dx$

Answer 1

$I=\int co{s}^2(4x+3)dx=\int \frac{1+cos2(4x+3)}{2}dx$

$[\because cos2\theta =2co{s}^2\theta -1]$

$=\int \frac{1+cos(8x+6)}{2}dx$

$=\frac{x}{2}+\frac{1}{8}\frac{sin(8x+6)}{2}+c$

$I=\frac{x}{2}+\frac{sin(8x+6)}{16}+c$