Example 39 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Integration by specific formulaes - Formula 3
Integration by specific formulaes - Formula 3
Last updated at April 16, 2024 by Teachoo
Example 39 Evaluate β«1β[β(cotβ‘π₯ )+β(tanβ‘π₯ )] ππ₯ β«1β[β(cotβ‘π₯ )+β(tanβ‘π₯ )] ππ₯ =β«1β[β(cotβ‘π₯ )+1/β(cotβ‘π₯ )] ππ₯ =β«1β[(cotβ‘π₯ + 1)/β(cotβ‘π₯ )] ππ₯ =β«1β[β(tanβ‘π₯ ) (cotβ‘π₯+1)] ππ₯ 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 β«1β[β(tanβ‘π₯ ) (cotβ‘π₯+1)] ππ₯ = β«1β[β(π‘^2 ) (cotβ‘π₯+1)] ππ₯ = β«1β[β(π‘^2 ) (1/tanβ‘π₯ +1)] ππ₯ = β«1βπ‘[1/π‘^2 +1] ππ₯ = β«1βπ‘[(1 + π‘^2)/π‘^2 ] Γ2π‘/(1 + π‘^4 ) . ππ‘ = β«1β2[(1 + π‘^2)/(1 + π‘^4 )] ππ‘ = 2β«1β(1 + π‘^2)/(1 + π‘^4 ) ππ‘ Dividing numerator and denominator by π‘^2 = 2 β«1β((1 + π‘^2)/π‘^2 )/((1 + π‘^4)/π‘^2 ) . ππ‘ = 2 β«1β(1/π‘^2 + 1)/(1/π‘^2 + π‘^2 ) . ππ‘ = 2 β«1β(1/π‘^2 + 1)/(1/π‘^2 + π‘^2 ) . ππ‘ = 2 β«1β(1 + 1/π‘^2 )/( π‘^2 + 1/π‘^2 + 2 β 2) . ππ‘ = 2 β«1β(1 + 1/π‘^2 )/( (π‘)^2 + (1/π‘)^2β 2 (π‘) (1/π‘) + 2) . ππ‘ = 2 β«1β(1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 + 2) . ππ‘ = 2 β«1β(1 + 1/π‘^2 )/((π‘ β 1/π‘)^2 +(β2 )^2 ) . ππ‘ 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 = 2 β«1β(1 + 1/π‘^2 )/(π¦^2 +(β2 )^2 ) . ππ‘ = 2 β«1β((1 + 1/π‘^2 ))/(π¦^2 + (β2 )^2 ) Γ ππ¦/((1 + 1/π‘^2 ) ) = 2 β«1β1/(π¦^2 + (β2 )^2 ) . ππ¦ = 2(1/β2 tan^(β1)β‘γ π¦/β2γ +πΆ1) = 2/β2 tan^(β1)β‘γ π¦/β2γ +2πΆ1 = β2 tan^(β1)β‘γ (1/π‘ β π‘)/β2γ +2πΆ1 = β2 tan^(β1)β‘γ (π‘^2 β 1)/(β2 π‘)γ +πΆ = β2 tan^(β1)β‘((tanβ‘π₯ β 1)/(β2 β(tanβ‘π₯ )))+πΆ = βπ γπππγ^(βπ)β‘((πππβ‘π β π)/(β(π πππβ‘π ) ))+πͺ