Question

Integrate:

$\int \cos x\log \cos xdx$

Answer 1

The equation of the value is,

$u=ln\left(cos\right(x\left)\right),$

$du=\frac{-sin\left(x\right)}{cos\left(x\right)}$

$dv=cos\left(x\right),$,

$v=sin\left(x\right)I$

Therefore $u\ast v-int(v\ast du)$$=ln\left(cos\right(x\left)\right)\ast sin\left(x\right)-sin\left(x\right)+ln\left(sec\right(x)+tan(x\left)\right)$

now evaluating at $\pi /2$ weget,

$=lncos\left(\frac{\pi }{2}\right)-1+lnsec\left(\frac{\pi }{2}\right)+lntan\left(\frac{\pi }{2}\right)$

$=\frac{lncos\left(\frac{\pi }{2}\right)-1+ln(1+sin)\left(\frac{\pi }{2}\right)}{\left(cos\frac{\pi }{2}\right)}$

$=ln(1+sin(\frac{\pi }{2}\left)\right)-1=ln\left(2\right)-1$