Integration by parts

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.6, 10 (sin^(β1)β‘π₯ )^2 β«1β(sin^(β1)β‘π₯ )^2 ππ₯ Let sin^(β1)β‘π₯=π β΄ π₯=sinβ‘π Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ=cosβ‘π ππ₯=cosβ‘π.ππ Thus, our equation becomes β«1β(sin^(β1)β‘π₯ )^2 ππ₯ = β«1βπ^2 . cosβ‘π.ππ =π^2 β«1βγcosβ‘π.ππγββ«1β(π(π^2 )/ππ β«1βγcosβ‘π.ππγ)ππ =π^2 sinβ‘πββ«1βγ2π.sinβ‘π. ππγ =π^2 sinβ‘πβ2β«1βγπ½ πππβ‘π½. ππ½γ 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) = ΞΈ2 and g(x) = cos ΞΈ Solving I1 β«1βγπ½ πππβ‘π½. ππ½γ = πβ«1βγsinβ‘π ππγββ«1β(ππ/ππ β«1βγsinβ‘π ππγ) ππ = π(βcosβ‘π )ββ«1βγ1.(βcosβ‘π ) γ ππ = βπ cosβ‘π+β«1βcosβ‘π ππ = (βπ cosβ‘π+sinβ‘π )+πΆ1 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) = ΞΈ and g(x) = sin ΞΈ Putting value of I1 in our equation β«1β(sin^(β1)β‘π₯ )^2 ππ₯ = π^2 sinβ‘πβ2β«1βγπ½ πππβ‘π½. ππ½γ =π^2 sinβ‘πβ2(βπ cosβ‘π+sinβ‘π+πΆ1) =π^2 sinβ‘π+2π cosβ‘πβ2 sinβ‘πβ2πΆ1 =π^2 sinβ‘π+2πβ(1βsin^2β‘γπ γ )β2 sinβ‘πβπͺπ =π^2 sinβ‘π+2πβ(1βsin^2β‘γπ γ )β2 sinβ‘π+πͺ =(γπ¬π’π§γ^(βπ)β‘π )^π π+π(γπ¬π’π§γ^(βπ)β‘π ) β(γπβγβ‘γπ^π γ )βππ+πͺ ππ πππ π=sin^(β1)β‘π₯ & sinβ‘π=π₯