1. Chapter 7 Class 12 Integrals (Term 2)
2. Concept wise
3. Integration by parts - e^x integration

Transcript

Ex 7.6, 18 Integrate the function - ππ₯ ((1 + sinβ‘π₯)/(1 + cosβ‘π₯ )) Simplifying function π^π₯ ((1 + sinβ‘π₯)/(1 + cosβ‘π₯ )) π^π₯ ((1 + sinβ‘π₯)/(1 + cosβ‘π₯ ))=π^π₯ ((1 + 2 sinβ‘(π₯/2) cosβ‘(π₯/2))/(2 γπππ ^2γβ‘(π₯/2) )) πππβ‘ππ=π πππβ‘π πππβ‘π Replacing x by π₯/2 , we get γπ ππ 2γβ‘(π₯/2)=2 π ππβ‘(π₯/2) πππ β‘(π₯/2) π ππβ‘π₯=2 π ππβ‘(π₯/2) πππ β‘(π₯/2) ππ¨π¬β‘ππ= 2γ"cos" γ^π π½βπ Replacing ππ₯ by π₯/2 γπππ  2γβ‘(π₯/2)=2πππ ^2 (π₯/2)β1 1+πππ β‘π₯=2πππ ^2 π₯/2 We know γπ ππ^2γβ‘π₯+γπππ ^2γβ‘π₯β‘= 1 Replacing x by π₯/2 , we get γπ ππ^2γβ‘(π₯/2)β‘γ+γπππ ^2γβ‘(π₯/2) γ=1 =π^π₯ ((γπππγ^πβ‘(π/π) + γπππγ^πβ‘(π/π) +2 γsin γβ‘γπ₯/2γ . γcos γβ‘γπ₯/2γ)/( 2 γπππ ^2γβ‘(π₯/2) )) =π^π₯ ((γsin γβ‘γπ₯/2γ + γcos γβ‘γπ₯/2γ )^2/( 2 γπππ ^2γβ‘(π₯/2) )) =π^π₯/2 ((γsin γβ‘γπ₯/2γ + γcos γβ‘γπ₯/2γ)/( πππ β‘γ π₯/2γ ))^2 =π^π₯/2 (γsin γβ‘γπ₯/2γ/( πππ β‘γ π₯/2γ ) + γcos γβ‘γπ₯/2γ/( πππ β‘γ π₯/2γ ))^2 =π^π₯/2 (γtan γβ‘γπ₯/2γ +1)^2 =π^π₯/2 (tan^2β‘γπ₯/2γ +(1)^2+2(γtan γβ‘γπ₯/2γ )(1)) =π^π₯/2 (γγπππγ^π γβ‘γπ/πγ +π+2 γtan γβ‘γπ₯/2γ ) =π^π₯/2 (γγπππγ^π γβ‘γπ/πγ +2 γtan γβ‘(π₯/2) ) =π^π₯ (1/2 . sec^2β‘γπ₯/2γ +2/2 γtan γβ‘(π₯/2) ) =π^π₯ (γtan γβ‘(π₯/2)+1/2 sec^2β‘(π₯/2) ) We know γ1+tan^2γβ‘π₯=sec^2β‘π₯ Replacing x by π₯/2 , we get γ1+γπ‘ππ^2γβ‘(π₯/2)γβ‘= sec^2 (π₯/2) Thus, Our Integration becomes β«1βγπ^π₯ " " ((1 + sinβ‘π₯)/(1 + cosβ‘π₯ )) ππ₯γ " "=β«1βγπ^π₯ (γtan γβ‘(π₯/2)+1/2 sec^2β‘(π₯/2) )ππ₯γ " " It is of the form β«1βγπ^π₯ [π(π₯)+π^β² (π₯)] γ ππ₯=π^π₯ π(π₯)+πΆ Where π(π₯)=tan^(β1)β‘π₯ π^β² (π₯)= 1/(1 + π₯^2 ) So, our equation becomes β«1βγπ^π₯ " " ((1 + sinβ‘π₯)/(1 + cosβ‘π₯ )) ππ₯γ " " = π^π γπππ γβ‘γπ/πγ+πͺ

Integration by parts - e^x integration

Chapter 7 Class 12 Integrals (Term 2)
Concept wise