Ex 7.3, 4 - Chapter 7 Class 12 Integrals
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 7.3, 4 sin3 (2 + 1) We know that sin 3 =3 sin 4 sin^3 4 sin^3 =3 sin sin 3 sin^3 =(3 sin sin 3 )/4 Replace by 2 +1 sin^3 (2 +1)=(3 sin (2 + 1) sin 3(2 + 1))/4 sin^3 (2 +1)=(3 sin (2 + 1) sin (6 + 3))/4 Thus, . 1 sin3 (2 +1) = 1 (3 sin (2 + 1) sin (6 + 3))/4 1 sin3 (2 +1) =1/4 1 (3 sin (2 +1) sin (6 +3) ) 1 sin3 (2 +1) =1/4 [3 1 sin (2 +1) 1 sin (6 +3) ] =1/4 [3 1/2 ( cos (2 +1) ) 1/6 ( cos (6 +3)+ )" " ] =1/4 [ 3/2 cos (2 +1)+1/6 cos (6 +3) ]+ = 3/8 cos (2 +1)+1/24 cos 3(2 +1)+ = 3/8 cos (2 +1)+1/24 [4 cos^3 (2 +1) 3 cos (2 +1) ]+ = 3/8 cos (2 +1)+4/24 cos^3 (2 +1) 3/24 cos (2 +1)+ = 3/8 cos (2 +1)+1/6 cos^3 (2 +1) 1/8 cos (2 +1)+ = 3/8 cos (2 +1) 1/8 cos (2 +1)+1/6 cos^3 (2 +1)+ = 4/8 cos (2 +1)+1/6 cos^3 (2 +1)+ = / ( + )+ / ^ ( + )+
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.