Question

10. 3 0(2log sinx - log sin 2x)dx

Answer 1

The integral is given as,

y= ∫ 0 π 2 ( 2logsinx−logsin2x ) dx

We have to calculate the integral of y.

Further we can simplify the integral as,

y= ∫ 0 π 2 ( 2logsinx−logsin2x ) dx = ∫ 0 π 2 ( 2logsinx−log( 2sinxcosx ) ) dx

Use the property ( logc⋅d )=log( c )+log( d ) to simplify the integral.

y= ∫ 0 π 2 ( 2 logsinx−log( 2 )−logsinx−logcosx ) dx = ∫ 0 π 2 ( logsinx−log( 2 )−logcosx ) dx = ∫ 0 π 2 ( log sinx ) dx− ∫ 0 π 2 ( log2 )  dx− ∫ 0 π 2 (  logcosx ) dx

We can apply the property ∫ 0 b f( x ) dx= ∫ 0 b f( b−x ) dxon the part ∫ 0 π 2 ( logcosx ) dx.

y =  ∫ 0 π 2 ( logsinx ) dx− ∫ 0 π 2 ( log2 ) dx− ∫ 0 π 2 ( logcos ( π 2 −x ) ) dx = ∫ 0 π 2 ( log sinx ) dx− ∫ 0 π 2 (  log2 ) dx− ∫ 0 π 2 (  logsinx ) dx =− ∫ 0 π 2 ( log 2 ) dx =−log2 [ x ] 0 π 2

Further simplify the above integral.

y=−log 2[ π 2 −0 ] =log ( 2 ) −1 [ π 2 ] = π 2 log( 1 2 )

Thus, the value of integral is π 2 log( 1 2 ).