Question

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

Answer 1

Let I=\(∫_0^π log(1+cosx)dx....(1)\)

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

\(I=∫_0^π log(1+cosx)dx....(2)\)

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

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

\(⇒2I=∫_0^π log(1-cos^2x)dx\)

\(⇒2I=∫_0^π logsin^2xdx\)

\(⇒2I=2∫_0^π logsinxdx\)

\(⇒I=∫_0^π logsinxdx...(3)\)

\(sin(π-x)=sinx\)

\(∴I=2∫_0^\frac{π}{2}logsinxdx...(4)\)

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

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

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

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

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

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

\(Let 2x=t  2dx=dt\)

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

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

\(⇒I=\frac{1πI}{2}-\frac{}{2}log2\)

\(⇒\frac{I}{2}=-\frac{π}{2}log2\)

\(⇒I=-πlog2\)