Question

By using the properties of definite integrals, evaluate the integral: $∫_0^π log(1+cosx)dx$

Answer 1

Let I=$∫_0^π log(1+cosx)dx....(1)$

$⇒I=∫_0^π log(1+cos(π-x))dx ......... (∫_0^aƒ(x)dx=∫_0^aƒ(a-x)dx)$

$I=∫_0^π log(1+cosx)dx....(2)$

$Adding(1)and(2),we obtain$

$2I=∫_0^π {log(1+cosx)+log(1-cosx)}dx$

$⇒2I=∫_0^π log(1-cos^2x)dx$

$⇒2I=∫_0^π logsin^2xdx$

$⇒2I=2∫_0^π logsinxdx$

$⇒I=∫_0^π logsinxdx...(3)$

$sin(π-x)=sinx$

$∴I=2∫_0^\frac{π}{2}logsinxdx...(4)$

$⇒I=2∫_0^\frac{π}{2} logsin(\frac{π}{2}-x)dx=2∫_0\frac{π}{2}logcosxdx...(5)$

$Adding(4)and(5),we obtain$

$2I=2∫_0^\frac{π}{2}(logsinx+logcosx)dx$

$⇒I=∫_0^\frac{π}{2}(logsinx+logcosx+log2-log2)dx$

$⇒I=∫_0^\frac{π}{2}(log2sinxcosx-log2)dx$

$⇒I=∫_0^\frac{π}{2}logsin2xdx-∫_0\frac{π}{2}log2dx$

$Let 2x=t 2dx=dt$

$When x=0,t=0 and when x=\frac{π}{2},π=$

$∴I=\frac{1π}{2}∫_0^π0logsintdt-\frac{}{2}log2$

$⇒I=\frac{1πI}{2}-\frac{}{2}log2$

$⇒\frac{I}{2}=-\frac{π}{2}log2$

$⇒I=-πlog2$