Ex 7.9

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.9, 2 Evaluate the integrals using substitution β«_0^(π/2)βγβ(sinβ‘γ" " Ο" " γ ) cos^5β‘Ο πΟγ Let πΌ=β«_0^(π/2)βγβπ ππΟ cos^5β‘γΟ πΟγ γ πΌ=β«_0^(π/2)βγβπ ππΟ cos^4β‘γΟ πππ Ο πΟγ γ πΌ=β«_0^(π/2)βγβ(sinβ‘Ο ) (1βsin^2β‘Ο )^2 πππ Ο πΟγ Put π‘=sinβ‘Ο Differentiating w.r.t. Ο ππ‘/πΟ=π(sinβ‘Ο )/πΟ ππ‘/πΟ=cosβ‘Ο ππ‘/cosβ‘Ο =πΟ Hence, when Ο varies form 0 to π/2 , π‘ varies form 0 to 1 Hence we can write the integral as πΌ=β«_0^(π/2)βγβπ ππΟ (1βsin^2β‘Ο )^2 πππ Ο πΟγ =β«_0^1βγβπ‘ (1βπ‘^2 )^2 πππ Ο ππ‘/πππ Ογ =β«_0^1βγβπ‘ (1βπ‘^2 )^2 ππ‘γ =β«_0^1βγπ‘^(1/2) (1βγ2π‘γ^2+π‘^4 )ππ‘γ =β«_0^1βγ (π‘^(1/2)βγ2π‘γ^(2 + 1/2)+π‘^(4 + 1/2) ) ππ‘γ =β«_0^1βγπ‘^(1/2) ππ‘γβ2β«1βπ‘^(3/2) ππ‘+β«1βπ‘^(9/2) ππ‘ =[π‘^(1/2 +1)/(1/2 +1)]_0^1β2[π‘^(3/2 +1)/(3/2 +1)]_0^1+[π‘^(9/2 +1)/(9/2 +1)]_0^1 =[π‘^(3/2 )/(3/2)]_0^1β2[π‘^(7/2 )/(7/2)]_0^1+[π‘^(11/2 )/(11/2)]_0^1 =2/3 (1^(3/2)β0^(3/2) )β2 Γ 2/7 (1^(7/2)β0^(7/2) )+2/11 [1^(11/2)β0^(11/2) ] =2/3β4/7+2/11 =(2 Γ 7 Γ 11 β 4 Γ 3 Γ 11 + 2 Γ 3 Γ7)/(2 Γ 7 Γ 11) =(154 β 132 + 42)/231 =ππ/πππ