Ex 7.2

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.2, 34 Integrate β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ Simplifying the function β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ = β(tanβ‘π₯ )/(sinβ‘γπ₯ cosβ‘π₯ γ. cosβ‘π₯/cosβ‘π₯ ) = β(tanβ‘π₯ )/(sinβ‘π₯ . cos^2β‘π₯/cosβ‘π₯ ) = β(tanβ‘π₯ )/(cos^2β‘π₯ . (sin π₯)/cosβ‘π₯ ) Concept: There are two methods to deal with π‘ππβ‘π₯ (1) Convert into π ππβ‘π₯ and πππ β‘π₯ , then solve using the properties of π ππβ‘π₯ and πππ β‘π₯ . (2) Change into sec2x, as derivative of tan x is sec2 . Here, 1st Method is not applicable , so we have used 2nd Method . = β(tanβ‘π₯ )/(cos^2β‘π₯ . tanβ‘π₯ ) = (tanβ‘π₯ )^(1/2 β 1) Γ 1/cos^2β‘π₯ = (tanβ‘π₯ )^((β1)/2) Γ 1/cos^2β‘π₯ = (tanβ‘π₯ )^((β1)/2) Γ sec^2β‘π₯ β΄ β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ " = " (tanβ‘π₯ )^((β1)/2) " Γ " sec^2β‘π₯ Step 2: Integrating the function β«1βγ β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ γ . ππ₯ = β«1βγ (tanβ‘π₯ )^((β1)/2) " Γ " sec^2β‘π₯ γ. ππ₯" " Let tanβ‘π₯ = π‘ Differentiating both sides π€.π.π‘.π₯ sec^2β‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/sec^2β‘π₯ Thus, our equation becomes β΄ β«1βγ (tanβ‘π₯ )^((β1)/2) " ." sec^2β‘π₯ γ. ππ₯" " = β«1βγ (π‘)^((β1)/2) " " . sec^2β‘π₯ γ. ππ‘/sec^2β‘π₯ " " = β«1βγπ‘^((β1)/2) . ππ‘γ = π‘^(β 1/2 +1)/(β 1/2 +1) + πΆ = π‘^(1/2)/(1/2) + πΆ = γ2π‘γ^(1/2)+ πΆ = 2βπ‘+ πΆ = πβ(π­ππ§β‘π )+ πͺ (Using π‘=π‘ππβ‘π₯)