Integration by parts

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.6, 13 Integrate the function - tan^(β1) π₯ β«1βγ" " tan^(β1) π₯" " γ .ππ₯=β«1βγ(tan^(β1) π₯) 1.ππ₯ " " γ = tan^(β1) π₯β«1βγ1 .γ ππ₯ββ«1β(π(tan^(β1)β‘π₯ )/ππ₯ β«1βγ1 .ππ₯γ) ππ₯ = tan^(β1) π₯ (π₯)ββ«1β1/(1 + π₯^2 ) . π₯ . ππ₯ = π₯ tan^(β1) π₯ββ«1βπ₯/(1 + π₯^2 ) . ππ₯ 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) = tanβ1 x and g(x) = 1 Solving I1 I1 = β«1βπ₯/(1 + π₯^2 ) . ππ₯" " Let 1 + π₯^2=π‘ Differentiating both sides π€.π.π‘.π₯ 0 + 2π₯=ππ‘/ππ₯ ππ₯=ππ‘/2π₯ Our equation becomes I1 = β«1βπ₯/(1 + π₯^2 ) . ππ₯" " Putting the value of (1+π₯^2 ) = t and ππ₯ = ππ‘/( 2π₯) , we get I1 = β«1βπ₯/π‘ . ππ‘/2π₯ I1 = 1/2 β«1β1/π‘ . ππ‘ I1 = 1/2 logβ‘γ |π‘|γ+πΆ1 I1 = 1/2 logβ‘γ |1+π₯^2 |γ+πΆ1 Putting the value of I1 in (1) , β«1βγ" " tan^(β1) π₯" " γ .ππ₯=π₯ tan^(β1) π₯ββ«1βπ₯/(1 + π₯^2 ) . ππ₯ =π₯ tan^(β1) π₯β(1/2 γlog γβ‘|1+π₯^2 |+πΆ1) =π₯ tan^(β1) π₯β1/2 γlog γβ‘|1+π₯^2 |βπΆ1 =π γπππγ^(βπ) πβπ/π γπππ γβ‘(π+π^π )+πͺ