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











Last updated at Dec. 20, 2019 by Teachoo
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Example 6 Find the following integrals: (i) β«1βγsin^3β‘π₯ cos^2β‘π₯ γ ππ₯ β«1βγsin^3β‘π₯ cos^2β‘π₯ γ ππ₯ Let cos π₯=π‘ Differentiating both sides π€.π.π‘.π₯. βsinβ‘π₯=ππ‘/ππ₯ ππ₯=(βππ‘)/sinβ‘π₯ Now are equation becomes β«1βγsin^3β‘π₯ cos^2β‘π₯ γ ππ₯ Putting value of πππ β‘π₯ and ππ₯ = β«1βsin^3β‘π₯ .π‘^2. ππ₯ = β«1βsin^3β‘π₯ .π‘^2. ππ‘/(βsinβ‘π₯ ) = β«1βsin^3β‘π₯/(βsinβ‘π₯ ) π‘^2. ππ‘ = ββ«1βsin^2β‘π₯ π‘^2. ππ‘ = β β«1β(1βcos^2β‘π₯ ) π‘^2. ππ‘ = β β«1β(1βπ‘^2 ) π‘^2. ππ‘ = β β«1β(π‘^2βπ‘^4 ) ππ‘ = β«1β(βπ‘^2+π‘^4 ) ππ‘ = β«1βγβπ‘^2 γ. ππ‘ + β«1βπ‘^4 . ππ‘ (β΄ sin^2β‘π₯=1βcos^2β‘π₯) = (γβπ‘γ^2+1)/(2 + 1)+π‘^(4 + 1)/(4 + 1)+πΆ = (βπ‘^3)/3 +π‘^5/5 +πΆ Putting back value of t = cos x = (βπ)/π γπππγ^πβ‘π +π/π γπππγ^πβ‘π +πͺ Example 6 Find the following integrals (ii) β«1βsinβ‘π₯/sinβ‘(π₯ + π) ππ₯ Let π₯+π=π‘ Differentiate both sides π€.π.π‘.π₯. 1=ππ‘/ππ₯ ππ₯=ππ‘ Hence, our equation becomes β«1βsinβ‘π₯/sinβ‘(π₯ + π) ππ₯ Putting the value of (π₯β‘+ π) and ππ₯ = β«1βsinβ‘π₯/sinβ‘π‘ ππ₯" " = β«1βπππβ‘(π β π)/sinβ‘π‘ ππ‘" " = β«1β(πππβ‘π ππ¨π¬β‘π β π¬π’π§β‘π ππ¨π¬β‘π)/sinβ‘π‘ ππ‘ = β«1β(sinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ β (sinβ‘π cosβ‘π‘)/sinβ‘π‘ ) ππ‘ = β«1βsinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ . ππ‘ββ«1β(sinβ‘π cosβ‘π‘)/sinβ‘π‘ . ππ‘ = β«1βcosβ‘π . ππ‘ββ«1βγsinβ‘π coπ‘β‘π‘ γ . ππ‘ = cos πβ«1β1. ππ‘βsinβ‘π β«1βπππβ‘π . π π = cos π (π‘)βsinβ‘π πππβ‘|πππβ‘π |+ πΆ = π‘.cosβ‘πβπ ππβ‘π πππβ‘|π ππβ‘π‘ |+ πΆ Putting back value of t = x + a (Using x + a = t β΄ x = t β a) (β(ππ πππ sinβ‘(πβπ)=@sinβ‘π cosβ‘πβsinβ‘π cosβ‘π )) ( β«1βcoπ‘β‘π₯ ππ₯=logβ‘|sinβ‘π₯ | ) = β«1βπππβ‘(π β π)/sinβ‘π‘ ππ‘" " = β«1βπππβ‘γπ ππ¨π¬β‘π βπ¬π’π§β‘π ππ¨π¬β‘π γ/sinβ‘π‘ ππ‘ = β«1β(sinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ β (sinβ‘π cosβ‘π‘)/sinβ‘π‘ ) ππ‘ = β«1βsinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ . ππ‘ββ«1β(sinβ‘π cosβ‘π‘)/sinβ‘π‘ . ππ‘ = β«1βcosβ‘π . ππ‘ββ«1βγsinβ‘π coπ‘β‘π‘ γ . ππ‘ = cos πβ«1β1. ππ‘βsinβ‘π β«1βπππβ‘π . π π = cos π (π‘)βsinβ‘π [πππβ‘|πππβ‘π | ]+πΆ1 = π‘.cosβ‘πβsinβ‘π [logβ‘|sinβ‘π‘ | ]+ πΆ1 Putting back value of t = x + a = (π₯+π) cosβ‘πβsinβ‘π [logβ‘|sinβ‘(π₯+π) | ]+ πΆ1 (Using x + a = t x = t β a) (β(ππ πππ sinβ‘(πβπ)=@sinβ‘π cosβ‘πβsinβ‘π cosβ‘π )) ( β«1βcoπ‘β‘π₯ ππ₯=logβ‘|sinβ‘π₯ | ) = (π₯+π) cosβ‘πβsinβ‘π logβ‘γ |sinβ‘(π₯+π) |γ+ πΆ = π₯ cosβ‘π+γπ cosγβ‘πβsinβ‘π logβ‘|sinβ‘(π₯+π) |+πΆ = π₯ cosβ‘πβsinβ‘π logβ‘|sinβ‘(π₯+π) |+γπ πππγβ‘π+πͺ = π πππβ‘πβπ¬π’π§β‘γ πγ πππβ‘|πππβ‘(π+π) |+ππ (πβπππ πΆ1=π cosβ‘π+πΆ) Example 6 Find the following integrals (iii) β«1β1/(1 + tanβ‘π₯ ) ππ₯ The given function cannot be integrated by direct substitution, so we have to simplify the given function . Simplifying the given function and integrating. β«1β1/(1 + tanβ‘π₯ ) .ππ₯ = β«1β1/(1 + sinβ‘π₯/cosβ‘π₯ ) .ππ₯ = β«1β1/((cosβ‘π₯ + sinβ‘π₯)/cosβ‘π₯ ) .ππ₯ (ππ πππ tanβ‘π₯=sinβ‘π₯/cosβ‘π₯ ) = β«1βcosβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ Multiplying and dividing by 2 = β«1β(2 cosβ‘π₯)/(2 (sinβ‘π₯ + cosβ‘π₯ ) ) .ππ₯ = β«1β(cosβ‘π₯ + cosβ‘π₯)/(2 (sinβ‘π₯ + cosβ‘π₯ ) ) .ππ₯ Adding and subtracting sinβ‘π₯ in the numerator = β«1β(cosβ‘π₯ + cosβ‘π₯ + sinβ‘π₯ β sinβ‘π₯)/(2 (sinβ‘π₯ + cosβ‘π₯ ) ) .ππ₯ = 1/2 β«1β(sinβ‘π₯ + cosβ‘π₯ + cosβ‘π₯ β sinβ‘π₯)/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ = 1/2 β«1β[(sinβ‘π₯ + cosβ‘π₯)/(sinβ‘π₯ + cosβ‘π₯ )+γcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ )] ππ₯ = 1/2 β«1β[1+γcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ )] ππ₯ = 1/2 [β«1βγ1.ππ₯+β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ )γ ππ₯] = 1/2 [π₯+πΆ1+β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) ππ₯] Take, I1 =β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ Let sinβ‘π₯ + cosβ‘π₯=π‘ Differentiate both sides π€.π.π‘.π₯. cosβ‘π₯βsinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) Putting value of (π ππβ‘π₯+πππ β‘π₯ ) and ππ₯ in I1 . I1=β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ I1 = β«1βγcosβ‘π₯ β sinγβ‘π₯/π‘ .ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) I1 =β«1β1/π‘ .ππ‘ Let sinβ‘π₯ + cosβ‘π₯=π‘ Differentiate both sides π€.π.π‘.π₯. cosβ‘π₯βsinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) Putting value of (π ππβ‘π₯+πππ β‘π₯ ) and ππ₯ in I1 . I1=β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ I1 = β«1βγcosβ‘π₯ β sinγβ‘π₯/π‘ .ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) I1 =β«1β1/π‘ .ππ‘ (ππ πππ β«1βγ1/π₯ .γ ππ₯=logβ‘γ |π₯|γ+πΆ) (ππ πππ π‘=sinβ‘π₯+cosβ‘π₯ ) (πβπππ πΆ=πΆ1/2+πΆ2/2)
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