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Integration by parts
Ex 7.6, 3
Ex 7.6, 23 (MCQ)
Example 17
Ex 7.6, 1
Ex 7.6, 2 Important
Ex 7.6, 12
Example 21 Important
Ex 7.6, 21
Ex 7.6, 5 Important
Ex 7.6, 4
Ex 7.6, 6
Ex 7.6, 15
Example 18 Important
Ex 7.6, 14 Important
Ex 7.6, 7 Important
Ex 7.6, 9
Ex 7.6, 8
Ex 7.6, 11
Example 20 Important
Ex 7.6, 13 Important You are here
Ex 7.6, 22 Important
Ex 7.6, 10 Important
Example 40 Important
Integration by parts
Last updated at Dec. 20, 2019 by Teachoo
Ex 7.6, 13 Integrate the function - tan^(β1) π₯ β«1βγ" " tan^(β1) π₯" " γ .ππ₯=β«1βγ(tan^(β1) π₯) 1.ππ₯ " " γ = tan^(β1) π₯β«1βγ1 .γ ππ₯ββ«1β(π(tan^(β1)β‘π₯ )/ππ₯ β«1βγ1 .ππ₯γ) ππ₯ = tan^(β1) π₯ (π₯)ββ«1β1/(1 + π₯^2 ) . π₯ . ππ₯ = π₯ tan^(β1) π₯ββ«1βπ₯/(1 + π₯^2 ) . ππ₯ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = tanβ1 x and g(x) = 1 Solving I1 I1 = β«1βπ₯/(1 + π₯^2 ) . ππ₯" " Let 1 + π₯^2=π‘ Differentiating both sides π€.π.π‘.π₯ 0 + 2π₯=ππ‘/ππ₯ ππ₯=ππ‘/2π₯ Our equation becomes I1 = β«1βπ₯/(1 + π₯^2 ) . ππ₯" " Putting the value of (1+π₯^2 ) = t and ππ₯ = ππ‘/( 2π₯) , we get I1 = β«1βπ₯/π‘ . ππ‘/2π₯ I1 = 1/2 β«1β1/π‘ . ππ‘ I1 = 1/2 logβ‘γ |π‘|γ+πΆ1 I1 = 1/2 logβ‘γ |1+π₯^2 |γ+πΆ1 Putting the value of I1 in (1) , β«1βγ" " tan^(β1) π₯" " γ .ππ₯=π₯ tan^(β1) π₯ββ«1βπ₯/(1 + π₯^2 ) . ππ₯ =π₯ tan^(β1) π₯β(1/2 γlog γβ‘|1+π₯^2 |+πΆ1) =π₯ tan^(β1) π₯β1/2 γlog γβ‘|1+π₯^2 |βπΆ1 =π γπππγ^(βπ) πβπ/π γπππ γβ‘(π+π^π )+πͺ