









Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Miscellaneous
Misc 2 Important
Misc 3 Important
Misc 4
Misc 5 Important
Misc 6
Misc 7 Important
Misc 8 Important
Misc 9
Misc 10 Important
Misc 11
Misc 12
Misc 13
Misc 14 Important
Misc 15
Misc 16
Misc 17
Misc 18 Important
Misc 19 Important
Misc 20 Important
Misc 21
Misc 22
Misc 23
Misc 24 Important
Misc 25 Important
Misc 26 Important
Misc 27 Important
Misc 28 Important You are here
Misc 29
Misc 30 Important
Misc 31 Important
Misc 32 Important
Misc 33 Important
Misc 34
Misc 35
Misc 36
Misc 37
Misc 38 Important
Misc 39
Misc 40 Important Deleted for CBSE Board 2023 Exams
Misc 41 (MCQ) Important
Misc 42 (MCQ)
Misc 43 (MCQ)
Misc 44 (MCQ) Important
Integration Formula Sheet - Chapter 7 Class 12 Formulas Important
Last updated at Dec. 23, 2019 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Misc 28 Evaluate the definite integral β«_(π/6)^(π/3)βγ(sinβ‘π₯ + cosβ‘π₯)/β(sinβ‘γ2π₯ γ ) γ β«_(π/6)^(π/3)βγ(sinβ‘π₯ + cosβ‘π₯)/β(sinβ‘γ2π₯ γ ) ππ₯ γ = β«_(π/6)^(π/3)βγ (sinβ‘π₯ + cosβ‘π₯)/β(2 sinβ‘π₯ cosβ‘π₯ ) ππ₯ γ = β«_(π/6)^(π/3)βγ (sinβ‘π₯/β(2 sinβ‘π₯ cosβ‘π₯ )+cosβ‘π₯/β(2 sinβ‘π₯ cosβ‘π₯ )) ππ₯ γ = β«_(π/6)^(π/3)βγ (β(sinβ‘π₯ )/(β2 β(cosβ‘π₯ ))+β(cosβ‘π₯ )/(β2 β(sinβ‘π₯ ))) ππ₯ γ ("Using" sinβ‘2π=2 sinβ‘π cosβ‘π ) = β«_(π/6)^(π/3)βγ (1/β2 β(sinβ‘π₯/cosβ‘π₯ ) + 1/β2 β(cosβ‘π₯/sinβ‘π₯ )) ππ₯ γ = 1/β2 β«_(π/6)^(π/3)βγ(β(tanβ‘π₯ )+β(cotβ‘π₯ ) ) ππ₯ γ = 1/β2 β«1_(π/6)^(π/3)βγ [β(cotβ‘π₯ )+1/β(cotβ‘π₯ )] γ ππ₯ = 1/β2 β«1_(π/6)^(π/3)β(cotβ‘π₯ + 1)/β(cotβ‘π₯ ) ππ₯ = 1/β2 β«1_(π/6)^(π/3)βγβ(tanβ‘π₯ ) (cotβ‘π₯+1) γ ππ₯ ("Using" π‘ππβ‘π₯= 1/πππ‘β‘π₯ ) Let tanβ‘π₯=π‘^2 Differentiating both sides π€.π.π‘.π₯. sec^2 π₯=2π‘ ππ‘/ππ₯ 1+tan^2 π₯=2π‘ . ππ‘/ππ₯ 1+(π‘^2 )^2=2π‘ . ππ‘/ππ₯ 1+π‘^4=2π‘ . ππ‘/ππ₯ (1+π‘^4 ) ππ₯=2π‘ ππ‘ ππ₯=2π‘/(1 + π‘^4 ) ππ‘ Putting values of t & dt, we get ("Using" π‘ππβ‘π₯= π‘^2) Putting values of t & dt, we get 1/β2 β«1_(π/6)^(π/3)βγβ(tanβ‘π₯ ) (cotβ‘π₯+1) γ ππ₯ =1/β2 β«1_(π/6)^(π/3)β[β(π‘^2 ) (cotβ‘π₯+1)] ππ₯ = 1/β2 β«1_(π/6)^(π/3)β[β(π‘^2 ) (1/tanβ‘π₯ +1)] ππ₯ = 1/β2 β«1_(π/6)^(π/3)βπ‘[1/π‘^2 +1] ππ₯ = 1/β2 β«1_(π/6)^(π/3)βπ‘[(1 + π‘^2)/π‘^2 ] ππ₯ = 1/β2 β«1_(π/6)^(π/3)βπ‘[(1 + π‘^2)/π‘^2 ] Γ2π‘/(1 + π‘^2 ) . ππ‘ = 1/β2 β«1_(π/6)^(π/3)β2[(1 + π‘^2)/(1 + π‘^4 )] ππ‘ = 1/β2 2β«1_(π/6)^(π/3)β(1 + π‘^2)/(1 + π‘^4 ) ππ‘ Dividing numerator and denominator by π‘^2 = β2 β«1_(π/6)^(π/3)βγ ((1 + π‘^2)/π‘^2 )/((1 + π‘^4)/π‘^2 )γ. ππ‘ = β2 β«1_(π/6)^(π/3)βγ (1/π‘^2 + 1)/(1/π‘^2 + π‘^2 )γ. ππ‘ = β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/( π‘^2 + 1/π‘^2 + 2 β 2). ππ‘ = β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/( (π‘)^2 + (1/π‘)^2β 2 (π‘) (1/π‘) + 2). ππ‘ = β2 β«1_(π/6)^(π/3)βγ (1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 + 2)γ. ππ‘ = β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 +(β2 )^2 ). ππ‘ (Adding and subtracting 2 in denominator) Let π‘β1/π‘=π¦ Differentiating both sides π€.π.π‘.π₯. 1+ 1/π‘^2 = ππ¦/ππ‘ ππ‘ =ππ¦/((1 + 1/π‘^2 ) ) Putting the values of (1/t βt) and dt, we get β2 β«1_(π/6)^(π/3)β(1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 +(β2 )^2 ). ππ‘ = β2 β«1_(π/6)^(π/3)βγ (1 + 1/π‘^2 )/(π¦^2 +(β2 )^2 )γ. ππ‘ = β2 β«1_(π/6)^(π/3)βγ ((1 + 1/π‘^2 ))/(π¦^2 +(β2 )^2 )γΓ ππ¦/((1 β 1/π‘^2 ) ) = β2 β«1_(π/6)^(π/3)βγ 1/(π¦^2 +(β2 )^2 )γ. ππ¦ = β2 (1/β2 tan^(β1)β‘γ π¦/β2γ )_(π/6)^(π/3) = (tan^(β1)β‘γ π¦/β2γ )_(π/6)^(π/3) = (tan^(β1)β‘γ (1/π‘ β π‘)/β2γ )_(π/6)^(π/3) ["Using" π¦=1/π‘ βπ‘] = (tan^(β1)β‘γ (π‘^2 β 1)/(β2 π‘)γ )_(π/6)^(π/3) = (tan^(β1)β‘((tanβ‘π₯ β 1)/(β2 β(tanβ‘π₯ ))) )_(π/6)^(π/3) = tan^(β1) ((tanβ‘(π/3) β 1)/β(2 tanβ‘(π/3) ))βtan^(β1) ((tanβ‘(π/6) β 1)/β(2 tanβ‘(π/6) )) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1/β3 β1 )/(β3 β(2 . 1/(β3 " " )))) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1 β β3 )/(β3 β(2 . 1/(β3 " " )))) Using π‘ππβ‘(π/3) = β3 π‘ππβ‘(π/6) = 1/(β3 " " ) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1 β β3 )/β(3 . 2 . 1/(β3 " " ))) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((1 β β3 )/β(2 β3) ) = tan^(β1) ((β3 β 1)/β(2 β3) )βtan^(β1) ((β(β3 β 1))/β(2 β3) ) = tan^(β1) ((β3 β 1)/β(2 β3) )+tan^(β1) ((β3 β 1)/β(2 β3) ) = 2 tan^(β1) ((β3 β 1)/β(2 β3) ) = π γπ¬π’πγ^(βπ) [(βπ β π)/π] "Using " tan^(β1)β‘(βπ) =βtan^(β1)β‘(π) Note π΄π΅^2=π΅πΆ^2+π΄πΆ^2 π΄π΅^2=(β3β1)^2+(β(2 β3) )^2 π΄π΅^2=3+1β2 β3+2 β3 π΄π΅^2=4 β΄ π΄π΅=2 sin π = π΅πΆ/π΄π΅ sin π = (β3 β 1)/2 π = γπ ππγ^(β1) ((β3 β 1)/2)