Question

integrala din
cos^4(x)

Answer 1

$\displaystyle \cos(2x)=2 \cos^2x-1 \Rightarrow \boxed{\cos^2x= \frac{1+ \cos(2x)}{2} }~. \\ \\ \cos^4x= \left( \cos ^2x \right)^2= \left( \frac{1+ \cos(2x)}{2}\right)^2= \frac{1+2 \cos(2x)+ \cos^2(2x)}{4}= \\ \\ = \frac{1+ 2 \cos(2x)+ \frac{1+ \cos(4x)}{2}}{4}= \frac{3+4 \cos(2x)+ \cos(4x)}{8}. \\ \\ \int \cos^4x \text{d} x= \int \left(\frac{3}{8}+ \frac{\cos(2x)}{2}+ \frac{\cos(4x)}{8} \right) \text{d}x = \\ \\ = \frac{3x}{8}+ \frac{\sin(2x)}{4}+ \frac{\sin(4x)}{32}+C.$