Ex 7.6

Chapter 7 Class 12 Integrals
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Ex 7.6, 22 Integrate the function sin^(β1) (2π₯/(1 + π₯2)) Simplifying the given function sin^(β1) (2π₯/(1 + π₯2)) Let π₯=tanβ‘π‘ β΄ π‘=tan^(β1)β‘(π₯) β΄ sin^(β1) (2π₯/(1 + π₯2))=sin^(β1) ((2 tanβ‘π‘" " )/(1 + tan^2β‘π‘ )) =sin^(β1) (sinβ‘2π‘ ) = 2t Ex 7.6, 22 Integrate the function sin^(β1) (2π₯/(1 + π₯2)) Simplifying the given function sin^(β1) (2π₯/(1 + π₯2)) Let π₯=tanβ‘π‘ β΄ π‘=tan^(β1)β‘(π₯) β΄ sin^(β1) (2π₯/(1 + π₯2))=sin^(β1) ((2 tanβ‘π‘" " )/(1 + tan^2β‘π‘ )) =sin^(β1) (sinβ‘2π‘ ) = 2t (ππ πππ sinβ‘2π=(2 tanβ‘π" " )/(1 + tan^2β‘π )) (ππ πππ sin^(β1) (sinβ‘π₯ )=π₯) =2 tan^(β1)β‘π₯ Thus, our function becomes β«1βγsin^(β1) (2π₯/(1 + π₯2)) ππ₯γ = 2β«1βγtan^(β1)β‘π₯ ππ₯γ =2β«1βγ(tan^(β1) π₯) 1.ππ₯ " " γ = 2 tan^(β1) π₯β«1βγ1 .γ ππ₯β2β«1β(π(tan^(β1)β‘π₯ )/ππ₯ β«1βγ1 .ππ₯γ) ππ₯ = 2tan^(β1) π₯ (π₯)β2β«1β1/(1 + π₯^2 ) . π₯ . 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 (ππ πππ π‘=tan^(β1)β‘(π₯) ) = 2π₯ tan^(β1) π₯β2β«1βπ₯/(1 + π₯^2 ) . ππ₯ 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sin^(β1) (2π₯/(1 + π₯2)) γ .ππ₯=2β«1βγ" " tan^(β1) π₯" " γ .ππ₯ =2π₯ tan^(β1) π₯β2β«1βπ₯/(1 + π₯^2 ) . ππ₯ =2π₯ tan^(β1) π₯β2(1/2 γlog γβ‘|1+π₯^2 |+πΆ1) (As t = 1 + x2) =2π₯ tan^(β1) π₯βγlog γβ‘|1+π₯^2 |β2πΆ1 =ππ γπππγ^(βπ) πβγπππ γβ‘(π+π^π )+πͺ As 1 + x2 is always positive |1+π₯^2 | = 1 + x2

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.