Ex 7.4, 20 - Integrate x + 2 / root 4x - x2 - Class 12 - Integration by specific formulaes - Method 10

Ex 7.4, 20 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 20 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.4, 20 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.4, 20 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.4, 20 - Chapter 7 Class 12 Integrals - Part 6

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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 = ( ^ )+ ^( ) (( )/ ) +

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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