Integration Full Chapter Explained - Integration Class 12 - Everything you need

1. Chapter 7 Class 12 Integrals
2. Concept wise
3. Definite Integration - By Formulae

Transcript

Ex 7.9, 8 β«_(π/6)^(π/4)βγπππ ππ π₯γβ‘ππ₯ Let F(π₯)=β«1βγπππ ππ π₯ . ππ₯γ Multiplying and Dividing by πππ ππ π₯+πππ‘ π₯ F(π₯)=β«1β(πππ ππ π₯ (πππ ππ π₯ + πππ‘ π₯))/(πππ ππ π₯ + πππ‘ π₯) ππ₯ Let cππ ππ π₯+πππ‘ π₯=π‘ Differentiating w.r.t. π₯ π/ππ₯ (πππ ππ π₯+πππ‘ π₯)=ππ‘/ππ₯ βπππ ππ^2 π₯βπππ ππ π₯ πππ‘ π₯=ππ‘/ππ₯ βπππ ππ π₯(πππ ππ π₯+πππ‘ π₯)=ππ‘/ππ₯ ππ₯=ππ‘/(βπππ ππ π₯ (πππ ππ π₯ + πππ‘ π₯) ) Therefore, β«1β(πππ ππ π₯(πππ ππ π₯ + πππ‘ π₯))/(πππ ππ π₯ + πππ‘ π₯) ππ₯ =β«1β(πππ ππ π₯(πππ ππ π₯ + πππ‘ π₯))/π‘. ππ‘/(βπππ ππ π₯(πππ ππ π₯ + πππ‘ π₯) ) =ββ«1βππ‘/π‘ =βlogβ‘γ |π‘|γ =βlogβ‘γ |πππ ππ π₯+πππ‘ π₯|γ Hence, F(π₯)=βlogβ‘|πππ ππ π₯+cotβ‘π₯ | Now, β«_(π/6)^(π/4)βγπππ ππ π₯=πΉ(π/4)βπΉ(π/6) γ =βπππ|πππ ππ(π/4)+πππ‘(π/4)|β (βlogβ‘|πππ ππ(π/6)+ πππ‘(π/6)| ) =βπππ|β2+1|+πππ|2+β3| =βπππ|2+β3|β πππ|β2+1| (Put π‘=πππ ππ π₯+πππ‘ π₯) =πππ|(2 + β3)/(β2 + 1)| =πππ|(2 + β3)/(β2 + 1)Γ(2 β β3)/(2 β β3)| =πππ[((2)^2 β (β3)^2)/((β2 + 1) Γ (2 β β3) )] =πππ[(4 β 3)/(β2 + 1)(2 β β3) ] =πππ[(1 Γ (β2 β 1))/((2 β β3)(β2 + 1) Γ(β2 β 1) )] =πππ[(β2 β 1)/(2 β β3)[(β2)^2β 1^2 ] ] =πππ[(βπ β π)/(π β βπ)] (log a β log b = log π/π)

Definite Integration - By Formulae