1. Chapter 7 Class 12 Integrals (Term 2)
2. Concept wise
3. Integration by substitution - Trignometric - Normal

Transcript

Example 6 Find the following integrals: (i) β«1βγsin^3β‘π₯ cos^2β‘π₯ γ ππ₯ β«1βγsin^3β‘π₯ cos^2β‘π₯ γ ππ₯ Let cos π₯=π‘ Differentiating both sides π€.π.π‘.π₯. βsinβ‘π₯=ππ‘/ππ₯ ππ₯=(βππ‘)/sinβ‘π₯ Now are equation becomes β«1βγsin^3β‘π₯ cos^2β‘π₯ γ ππ₯ Putting value of πππ β‘π₯ and ππ₯ = β«1βsin^3β‘π₯ .π‘^2. ππ₯ = β«1βsin^3β‘π₯ .π‘^2. ππ‘/(βsinβ‘π₯ ) = β«1βsin^3β‘π₯/(βsinβ‘π₯ ) π‘^2. ππ‘ = ββ«1βsin^2β‘π₯ π‘^2. ππ‘ = β β«1β(1βcos^2β‘π₯ ) π‘^2. ππ‘ = β β«1β(1βπ‘^2 ) π‘^2. ππ‘ = β β«1β(π‘^2βπ‘^4 ) ππ‘ = β«1β(βπ‘^2+π‘^4 ) ππ‘ = β«1βγβπ‘^2 γ. ππ‘ + β«1βπ‘^4 . ππ‘ (β΄ sin^2β‘π₯=1βcos^2β‘π₯) = (γβπ‘γ^2+1)/(2 + 1)+π‘^(4 + 1)/(4 + 1)+πΆ = (βπ‘^3)/3 +π‘^5/5 +πΆ Putting back value of t = cos x = (βπ)/π γπππγ^πβ‘π +π/π γπππγ^πβ‘π +πͺ