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Ex 7.4, 20 ( + 2)/ (4 ^2 ) 1 ( + 2)/ (4 ^2 )= 1/2 1 ( 2 4)/ (4 ^2 ) = 1/2 1 ( 2 + 4 4 4)/ (4 ^2 ) = 1/2 1 ( 2 + 4 8)/ (4 ^2 ) = 1/2 1 ( 2 + 4)/ (4 ^2 ) +8/2 1 / (4 ^2 ) = 1/2 1 ( 2 + 4)/ (4 ^2 ) +4 1 / (4 ^2 ) Solving I1= 1/2 1 (4 2 )/ (4 ^2 ) . Let 4 ^2= Diff both sides w.r.t. x 4 2 = / = /(4 2 ) Thus, our equation becomes I1= 1/2 1 (4 2 )/ (4 ^2 ) . Putting the values of (4 ^2 ) and I1= 1/2 1 (4 2 )/ . I1= 1/2 1 (4 2 )/ . /(4 2 ) I1= 1/2 1 1/ I1= 1/2 1 1/( )^(1/2) I1= 1/2 ^(( 1)/2 + 1)/(( 1)/2 + 1) + 1 I1= 1/2 ( ^(1/2 ))/(1/2) + 1 I1= ^(1/2 )+ 1 I1= + 1 I1= (4 ^2 )+ 1 Solving I2=4 1 1/ (4 ^2 ) . I2=4 1 1/ ( ( ^2 4 ) ) . I2=4 1 1/ ( ( ^2 2(2)( )) ) . I2=4 1 1/ ( ( ^2 2(2)( ) + (2)^2 (2)^2 ) ) . I2=4 1 1/ ( [( 2)^2 (2)^2 ] ) . I2=4 1 1/ ( [( 2)^2 4] ) . I2=4 1 1/ (4 ( 2)^2 ) . I2=4 1 1/ ((2)^2 ( 2)^2 ) . I2=4 1 1/ ((2)^2 ( 2)^2 ) . I2=4[ sin^( 1) (( 2)/2) ]+ 2 Putting values of I1 and I2 in eq. (1) 1 ( + 2)/ (4 ^2 ) . = 1/2 1 (4 2 )/ (4 ^2 ) . +4 1 1" " / (4 ^2 ) . = (4 ^2 )+ 1+4 sin^( 1) (( 2)/2) + 2 = ( ^ )+ ^( ) (( )/ ) +

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo