Ex 7.6, 22 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
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Ex 7.6, 23 (MCQ)
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Last updated at April 16, 2024 by Teachoo
Ex 7.6, 22 Integrate the function sin^(β1) (2π₯/(1 + π₯2)) Simplifying the given function sin^(β1) (2π₯/(1 + π₯2)) Let π₯=tanβ‘π‘ β΄ π‘=tan^(β1)β‘(π₯) β΄ sin^(β1) (2π₯/(1 + π₯2))=sin^(β1) ((2 tanβ‘π‘" " )/(1 + tan^2β‘π‘ )) =sin^(β1) (sinβ‘2π‘ ) = 2t Ex 7.6, 22 Integrate the function sin^(β1) (2π₯/(1 + π₯2)) Simplifying the given function sin^(β1) (2π₯/(1 + π₯2)) Let π₯=tanβ‘π‘ β΄ π‘=tan^(β1)β‘(π₯) β΄ sin^(β1) (2π₯/(1 + π₯2))=sin^(β1) ((2 tanβ‘π‘" " )/(1 + tan^2β‘π‘ )) =sin^(β1) (sinβ‘2π‘ ) = 2t (ππ πππ sinβ‘2π=(2 tanβ‘π" " )/(1 + tan^2β‘π )) (ππ πππ sin^(β1) (sinβ‘π₯ )=π₯) =2 tan^(β1)β‘π₯ Thus, our function becomes β«1βγsin^(β1) (2π₯/(1 + π₯2)) ππ₯γ = 2β«1βγtan^(β1)β‘π₯ ππ₯γ =2β«1βγ(tan^(β1) π₯) 1.ππ₯ " " γ = 2 tan^(β1) π₯β«1βγ1 .γ ππ₯β2β«1β(π(tan^(β1)β‘π₯ )/ππ₯ β«1βγ1 .ππ₯γ) ππ₯ = 2tan^(β1) π₯ (π₯)β2β«1β1/(1 + π₯^2 ) . π₯ . 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 (ππ πππ π‘=tan^(β1)β‘(π₯) ) = 2π₯ tan^(β1) π₯β2β«1βπ₯/(1 + π₯^2 ) . ππ₯ 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«1βγsin^(β1) (2π₯/(1 + π₯2)) γ .ππ₯=2β«1βγ" " tan^(β1) π₯" " γ .ππ₯ =2π₯ tan^(β1) π₯β2β«1βπ₯/(1 + π₯^2 ) . ππ₯ =2π₯ tan^(β1) π₯β2(1/2 γlog γβ‘|1+π₯^2 |+πΆ1) (As t = 1 + x2) =2π₯ tan^(β1) π₯βγlog γβ‘|1+π₯^2 |β2πΆ1 =ππ γπππγ^(βπ) πβγπππ γβ‘(π+π^π )+πͺ As 1 + x2 is always positive |1+π₯^2 | = 1 + x2