Ex 7.6

Chapter 7 Class 12 Integrals
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Ex 7.6, 7 (Method 1) π₯ sin^(β1)β‘π₯ β«1βγπ₯ γπ ππγ^(β1) γ π₯ ππ₯ Let x = sinβ‘π dx = cosβ‘π ππ Substituting values, we get β«1βγπ₯ γπ ππγ^(β1) γ π₯ ππ₯ = β«1βγsinβ‘π γπππγ^(βπ)β‘(πππβ‘π½ ) cosβ‘π ππ" " γ = β«1βγsinβ‘π π½ cosβ‘π ππ" " γ = 1/2 β«1βγπ γ(2 sinγβ‘π cosβ‘γπ)γ ππ" " γ = 1/2 β«1βγγπ sinγβ‘2π ππγ (β΅sin^(β1)β‘γ(sinβ‘π)γ = π) (β΅2β‘sinβ‘π₯ cosβ‘π₯ = β‘sinβ‘2π₯ ) Integrating by parts β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = ΞΈ and g(x) = sin 2ΞΈ = 1/2 [πβ«sinβ‘2π ππββ«(ππ/ππβ«sin 2π ππ)ππ] = 1/2 [π (β(cosβ‘2π))/2ββ«1.(βγ(cosγβ‘2π))/2 ππ] = 1/2 [(βπ)/2 cosβ‘2π+1/2β«cos 2π ππ] = 1/2 [(βπ)/2 cosβ‘2π+1/2 ( sinβ‘2π)/2] + C = (βπ)/4 πππβ‘ππ½ + 1/8 πππβ‘ππ½ + C = (βπ)/4 (1 β γππππγ^π π½) + 1/8 Γ 2 πππβ‘π½ ππ¨π¬ π½ + C = (βπ)/4 (1 β γ2π ππγ^2 π) + 1/4 π ππβ‘π πππ  π + C = (βπ)/4 (1 β γ2π ππγ^2 π) + 1/4 π ππβ‘π β(1βsin^2 π) + C = (βsin^(β1)β‘π₯)/4 (1 β 2π₯^2 )+π₯/4 β(1βπ₯^2 )+C = ((γππγ^π β π))/π γπππγ^(βπ) π+π/π β(πβπ^π )+π Using π₯=sinβ‘π βΉ π=sin^(β1)β‘π₯ Ex 7.6, 7 (Method 2) π₯ sin^(β1)β‘π₯ β«1βγπ₯ sin^(β1)β‘π₯ γ ππ₯ =sin^(β1)β‘π₯ β«1βπ₯ ππ₯ββ«1β(π(sin^(β1)β‘π₯ )/ππ₯ β«1βγπ₯ .ππ₯γ)ππ₯ = sin^(β1)β‘π₯. π₯^2/2 ββ«1βγ1/β(1 β π₯^2 ) . π₯^2/2. ππ₯γ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x and g(x) = sinβ1 x = π₯^2/2 sin^(β1)β‘π₯β1/2 β«1βγπ₯^2/β(1 β π₯^2 ) ππ₯γ Solving I1 I1 = 1/2 β«1βγπ₯^2/β(1 β π₯^2 ) ππ₯γ Multiplying and dividing by β1. I1 = 1/2 β«1βγγβπ₯γ^2/( ββ(1 β π₯^2 )) ππ₯γ I1 = (β1)/2 β«1βγγβπ₯γ^2/( β(1 β π₯^2 )) ππ₯γ I1 =(β1)/2 β«1βγγβ1 + 1 β π₯γ^2/( β(1 β π₯^2 )) ππ₯γ ("Adding and subtracting" "1 in numerator" ) I1 = (β1)/2 β«1βγ((β1)/β(1 β π₯^2 )+(1 βγ π₯γ^2)/β(1 β π₯^2 )) ππ₯γ I1 = (β1)/2 (β«1βγ(βπ)/β((π)^π βπ^π ).ππγ+β«1β(1βπ₯^2 )^(1 β 1/2) . ππ₯) I1 = (β1)/2 (βγπππγ^(βπ)β‘π +β«1β(1βπ₯^2 )^(1/2) . ππ₯) I1 = (β1)/2 (βsin^(β1)β‘π₯ +β«1ββ((1)^2βπ₯^2 ). ππ₯) I1 = (β1)/2 (βsin^(β1)β‘π₯ +π₯/2 β(1βπ₯^2 )+(1)^2/2 sin^(β1)β‘γ π₯/1γ+πΆ1) Using β«1βγππ₯/β(π^2 β π₯^2 )=sin^(β1)β‘γπ₯/πγ +πΆγ ππ πππ β(π^2βπ₯^2 ) ππ₯=1/π₯ π₯β(π^2βπ₯^2 )+π^2/2 sin^(β1)β‘γπ₯/πγ+πΆ I1 = (β1)/2 (βsin^(β1)β‘π₯+1/2 γ sinγ^(β1)β‘π₯+π₯/2 β(1βπ₯^2 )β‘+ πΆ1) I1 = (β1)/2 ((βsin^(β1)β‘π₯)/2 +π₯/2 β(1βπ₯^2 )β‘+ πΆ1) I1 = sin^(β1)β‘π₯/4 β π₯/4 β(1βπ₯^2 )β‘β πΆ1/2 Putting the value of I1 in eq. (1), we get β«1βγπ₯ sin^(β1)β‘π₯ ππ₯γ=π₯^2/2 sin^(β1)β‘π₯β(sin^(β1)β‘π₯/4 β π₯/4 β(1βπ₯^2 )β‘β πΆ1/2) =π₯^2/2 sin^(β1)β‘π₯βsin^(β1)β‘π₯/4+ π₯/4 β(1βπ₯^2 )β‘+ πΆ1/2 = sin^(β1)β‘π₯ (π₯^2/2β1/4)+ π₯/4 β(1βπ₯^2 )β‘+ πΆ = sin^(β1)β‘π₯ ((γ2π₯γ^2 β1)/4)+ π₯/4 β(1βπ₯^2 )β‘+ πΆ = ((γππγ^π βπ)/π) γπππγ^(βπ)β‘π+ π/π β(πβπ^π )β‘+ πͺ