Miscellaneous

Chapter 7 Class 12 Integrals
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Misc 27 Evaluate the definite integral β«_(π/6)^(π/3)βγ(sinβ‘π₯ + cosβ‘π₯)/β(sinβ‘γ2π₯ γ ) γ β«_(π/6)^(π/3)βγ(sinβ‘π₯ + cosβ‘π₯)/β(sinβ‘γ2π₯ γ ) ππ₯ γ = β«_(π/6)^(π/3)βγ (sinβ‘π₯ + cosβ‘π₯)/β(2 sinβ‘π₯ cosβ‘π₯ ) ππ₯ γ = β«_(π/6)^(π/3)βγ (sinβ‘π₯/β(2 sinβ‘π₯ cosβ‘π₯ )+cosβ‘π₯/β(2 sinβ‘π₯ cosβ‘π₯ )) ππ₯ γ = β«_(π/6)^(π/3)βγ (β(sinβ‘π₯ )/(β2 β(cosβ‘π₯ ))+β(cosβ‘π₯ )/(β2 β(sinβ‘π₯ ))) ππ₯ γ ("Using" sinβ‘2π=2 sinβ‘π cosβ‘π ) = β«_(π/6)^(π/3)βγ (1/β2 β(sinβ‘π₯/cosβ‘π₯ ) + 1/β2 β(cosβ‘π₯/sinβ‘π₯ )) ππ₯ γ = 1/β2 β«_(π/6)^(π/3)βγ(β(tanβ‘π₯ )+β(cotβ‘π₯ ) ) ππ₯ γ = 1/β2 β«1_(π/6)^(π/3)βγ [β(cotβ‘π₯ )+1/β(cotβ‘π₯ )] γ ππ₯ = 1/β2 β«1_(π/6)^(π/3)β(cotβ‘π₯ + 1)/β(cotβ‘π₯ ) ππ₯ = 1/β2 β«1_(π/6)^(π/3)βγβ(tanβ‘π₯ ) (cotβ‘π₯+1) γ ππ₯ ("Using" π‘ππβ‘π₯= 1/πππ‘β‘π₯ ) Let tanβ‘π₯=π‘^2 Differentiating both sides π€.π.π‘.π₯. sec^2 π₯=2π‘ ππ‘/ππ₯ 1+tan^2 π₯=2π‘ . ππ‘/ππ₯ 1+(π‘^2 )^2=2π‘ . ππ‘/ππ₯ 1+π‘^4=2π‘ . ππ‘/ππ₯ (1+π‘^4 ) ππ₯=2π‘ ππ‘ ππ₯=2π‘/(1 + π‘^4 ) ππ‘ Putting values of t & dt, we get ("Using" π‘ππβ‘π₯= π‘^2) Putting values of t & dt, we get 1/β2 β«1_(π/6)^(π/3)βγβ(tanβ‘π₯ ) (cotβ‘π₯+1) γ ππ₯ =1/β2 β«1_(π/6)^(π/3)β[β(π‘^2 ) (cotβ‘π₯+1)] ππ₯ = 1/β2 β«1_(π/6)^(π/3)β[β(π‘^2 ) (1/tanβ‘π₯ +1)] ππ₯ = 1/β2 β«1_(π/6)^(π/3)βπ‘[1/π‘^2 +1] ππ₯ = 1/β2 β«1_(π/6)^(π/3)βπ‘[(1 + π‘^2)/π‘^2 ] ππ₯ = 1/β2 β«1_(π/6)^(π/3)βπ‘[(1 + π‘^2)/π‘^2 ] Γ2π‘/(1 + π‘^2 ) . ππ‘ = 1/β2 β«1_(π/6)^(π/3)β2[(1 + π‘^2)/(1 + π‘^4 )] ππ‘ = 1/β2 2β«1_(π/6)^(π/3)β(1 + π‘^2)/(1 + π‘^4 ) ππ‘ Dividing numerator and denominator by π‘^2 = β2 β«1_(π/6)^(π/3)βγ ((1 + π‘^2)/π‘^2 )/((1 + π‘^4)/π‘^2 )γ. ππ‘ = β2 β«1_(π/6)^(π/3)βγ (1/π‘^2 + 1)/(1/π‘^2 + π‘^2 )γ. ππ‘ = β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/( π‘^2 + 1/π‘^2 + 2 β 2). ππ‘ = β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/( (π‘)^2 + (1/π‘)^2β 2 (π‘) (1/π‘) + 2). ππ‘ = β2 β«1_(π/6)^(π/3)βγ (1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 + 2)γ. ππ‘ = β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 +(β2 )^2 ). ππ‘ (Adding and subtracting 2 in denominator) Let π‘β1/π‘=π¦ Differentiating both sides π€.π.π‘.π₯. 1+ 1/π‘^2 = ππ¦/ππ‘ ππ‘ =ππ¦/((1 + 1/π‘^2 ) ) Putting the values of (1/t βt) and dt, we get β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 +(β2 )^2 ). ππ‘ = β2 β«1_(π/6)^(π/3)βγ (1 + 1/π‘^2 )/(π¦^2 +(β2 )^2 )γ. ππ‘ = β2 β«1_(π/6)^(π/3)βγ ((1 + 1/π‘^2 ))/(π¦^2 +(β2 )^2 )γΓ ππ¦/((1 β 1/π‘^2 ) ) = β2 β«1_(π/6)^(π/3)βγ 1/(π¦^2 +(β2 )^2 )γ. ππ¦ = β2 (1/β2 tan^(β1)β‘γ π¦/β2γ )_(π/6)^(π/3) = (tan^(β1)β‘γ π¦/β2γ )_(π/6)^(π/3) = (tan^(β1)β‘γ (1/π‘ β π‘)/β2γ )_(π/6)^(π/3) ["Using" π¦=1/π‘ βπ‘] = (tan^(β1)β‘γ (π‘^2 β 1)/(β2 π‘)γ )_(π/6)^(π/3) = (tan^(β1)β‘((tanβ‘π₯ β 1)/(β2 β(tanβ‘π₯ ))) )_(π/6)^(π/3) = tan^(β1) ((tanβ‘(π/3) β 1)/β(2 tanβ‘(π/3) ))βtan^(β1) ((tanβ‘(π/6) β 1)/β(2 tanβ‘(π/6) )) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1/β3 β1 )/(β3 β(2 . 1/(β3 " " )))) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1 β β3 )/(β3 β(2 . 1/(β3 " " )))) Using π‘ππβ‘(π/3) = β3 π‘ππβ‘(π/6) = 1/(β3 " " ) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1 β β3 )/β(3 . 2 . 1/(β3 " " ))) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1 β β3 )/β(2 β3) ) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((β(β3 β 1))/β(2 β3) ) = tan^(β1) ((β3 β 1)/β(2 β3) )+tan^(β1) ((β3 β 1)/β(2 β3) ) = 2 tan^(β1) ((β3 β 1)/β(2 β3) ) = π γπ¬π’πγ^(βπ) [(βπ β π)/π] "Using " tan^(β1)β‘(βπ) =βtan^(β1)β‘(π) Note π΄π΅^2=π΅πΆ^2+π΄πΆ^2 π΄π΅^2=(β3β1)^2+(β(2 β3) )^2 π΄π΅^2=3+1β2 β3+2 β3 π΄π΅^2=4 β΄ π΄π΅=2 sin π = π΅πΆ/π΄π΅ sin π = (β3 β 1)/2 π = γπ ππγ^(β1) ((β3 β 1)/2)

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#### 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.