Question

By using the properties of definite integrals, evaluate the integral: \(∫^\frac{π}{2}_0(2logsinx-logsin2x)dx\)

Answer 1

Let I=\(∫^\frac{π}{2}_0(2logsinx-logsin2x)dx\)

\(⇒I=∫^\frac{π}{2}_0{{2log sinx-log(2sinx cosx)}}dx\)

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

\(⇒I=∫^{π}{2}_0{{logsinx-logcosx-log2}}dx...(1)\)

\(It is known that,(∫^a_0ƒ(x)dx=∫^a_0ƒ(a-x)dx)\)

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

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

\(2I=∫^{π}{2}_0(-log2-log2)dx\)

\(⇒2I=-2log2∫^{π}{2}_01.dx\)

\(⇒I=-log2[\frac{π}{2}]\)

\(⇒I=\frac{π}{2}(-log2)\)

\(⇒I=\frac{π}{2}[log\frac{1}{2}]\)

\(⇒I=\frac{π}{2}log\frac{1}{2}\)