Integration by parts

Chapter 7 Class 12 Integrals
Concept wise

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

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Ex 7.6, 9 (Method 1) π₯ γcππ ^(β1)γβ‘π₯ β«1βπ₯ cos^(β1)β‘γπ₯ ππ₯γ Let x = cosβ‘π dx = β sinβ‘γπ ππγ Substituting values, we get β«1βπ₯ cos^(β1)β‘γπ₯ ππ₯ γ = β«1βcosβ‘π γπππγ^(βπ) (πππβ‘π½)(βsinβ‘γπ )ππγ = ββ«1βγγ sinγβ‘γπ γ π½ cosβ‘π ππγ = (β1)/2 β«1βγπ (2sinγβ‘γπ cosβ‘γπ) γ ππγ = (β1)/2 β«1βγγπ sinγβ‘2π ππγ Ex 7.6, 9 (Method 1) π₯ γcππ ^(β1)γβ‘π₯ β«1βπ₯ cos^(β1)β‘γπ₯ ππ₯γ Let x = cosβ‘π dx = β sinβ‘γπ ππγ Substituting values, we get β«1βπ₯ cos^(β1)β‘γπ₯ ππ₯ γ = β«1βcosβ‘π γπππγ^(βπ) (πππβ‘π½)(βsinβ‘γπ )ππγ = ββ«1βγγ sinγβ‘γπ γ π½ cosβ‘π ππγ = (β1)/2 β«1βγπ (2sinγβ‘γπ cosβ‘γπ) γ ππγ = (β1)/2 β«1βγγπ sinγβ‘2π ππγ (β΅γπππ γ^(β1)β‘γ(πππ β‘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 = (β1)/2 [πβ«sinβ‘2π ππββ«(ππ/ππβ«sin 2π ππ)ππ] = (β1)/2 [π (β(cosβ‘2π))/2ββ«1.(βγ(cosγβ‘2π))/2 ππ] = (β1)/2 [(βπ)/2 cosβ‘2π+1/2β«cos 2π ππ] = (β1)/2 [(βπ)/2 cosβ‘2π+1/2 ( sinβ‘2π)/2] + C = π/4 πππβ‘ππ½ β 1/8 πππβ‘ππ½ + C = π/4 (γππππγ^π π½ β 1) β 1/8 Γ 2 πππβ‘π½ ππ¨π¬ π½ + C = π/4 (γ2πππ γ^2 π β 1) β 1/4 π ππβ‘π πππ  π + C = π/4 (γ2πππ γ^2 π β 1) β 1/4 πππ β‘π β(1βcos^2 π) + C = cos^(β1)β‘π₯/4 (2π₯^2β1)βπ₯/4 β(1βπ₯^2 )+C = ((γππγ^π β π))/π γπππγ^(βπ) πβπ/π β(πβπ^π )+π Using π₯=cosβ‘π βΉ π=cos^(β1)β‘π₯ Ex 7.6, 9 (Method 2) π₯ cos^(β1)β‘π₯ β«1βγπ₯ cos^(β1)β‘π₯ γ ππ₯ =cos^(β1)β‘π₯ β«1βπ₯ ππ₯ββ«1β(π(cos^(β1)β‘π₯ )/ππ₯ β«1βγπ₯ .ππ₯γ)ππ₯ = cos^(β1)β‘π₯. π₯^2/2 ββ«1βγ(β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) = cosβ1 x = π₯^2/2 cos^(β1)β‘π₯+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 β¦(1) ("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₯ cos^(β1)β‘π₯ ππ₯γ=π₯^2/2 cos^(β1)β‘π₯+(sin^(β1)β‘π₯/4 β π₯/4 β(1βπ₯^2 )β‘β πΆ1/2) =π₯^2/2 cos^(β1)β‘π₯+1/4 sin^(β1)β‘π₯β π₯/4 β(1βπ₯^2 )β‘+ πΆ1/2 =π₯^2/2 cos^(β1)β‘π₯+1/4 (π/2βcos^(β1)β‘π₯ )β π₯/4 β(1βπ₯^2 )β‘+ πΆ1/2 =π₯^2/2 cos^(β1)β‘π₯+π/8β1/4 cos^(β1)β‘π₯β π₯/4 β(1βπ₯^2 )β‘+ πΆ1/2 =π₯^2/2 cos^(β1)β‘π₯β1/4 cos^(β1)β‘π₯β π₯/4 β(1βπ₯^2 )β‘+ πΆ1/2 +π/8 =π₯^2/2 cos^(β1)β‘π₯β1/4 cos^(β1)β‘π₯β π₯/4 β(1βπ₯^2 )β‘+ πΆ = cos^(β1)β‘π₯ (π₯^2/2β1/4)β π₯/4 β(1βπ₯^2 )β‘+ πΆ = cos^(β1)β‘π₯ ((γ2π₯γ^2 β 1)/4)β π₯/4 β(1βπ₯^2 )β‘+ πΆ = ((γππγ^π+ π)/π) γπππγ^(βπ)β‘πβ π/π β(πβπ^π )β‘+ πͺ