


Something went wrong!
The
video
couldn't load due to a technical hiccup.
But don't worry — our team is already on it, and we're working hard to get it back up ASAP.
Thanks for bearing with us!
Ex 7.2
Last updated at Dec. 16, 2024 by Teachoo
Something went wrong!
The
video
couldn't load due to a technical hiccup.
But don't worry — our team is already on it, and we're working hard to get it back up ASAP.
Thanks for bearing with us!
Transcript
Ex 7.2, 34 Integrate β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ Simplifying the function β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ = β(tanβ‘π₯ )/(sinβ‘γπ₯ 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‘ππβ‘π₯ (1) Convert into π ππβ‘π₯ and πππ β‘π₯ , then solve using the properties of π ππβ‘π₯ and πππ 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₯ )^((β1)/2) Γ 1/cos^2β‘π₯ = (tanβ‘π₯ )^((β1)/2) Γ sec^2β‘π₯ β΄ β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ " = " (tanβ‘π₯ )^((β1)/2) " Γ " sec^2β‘π₯ Step 2: Integrating the function β«1βγ β(tanβ‘π₯ )/sinβ‘γπ₯ cosβ‘π₯ γ γ . ππ₯ = β«1βγ (tanβ‘π₯ )^((β1)/2) " Γ " sec^2β‘π₯ γ. ππ₯" " Let tanβ‘π₯ = π‘ Differentiating both sides π€.π.π‘.π₯ sec^2β‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/sec^2β‘π₯ Thus, our equation becomes β΄ β«1βγ (tanβ‘π₯ )^((β1)/2) " ." sec^2β‘π₯ γ. ππ₯" " = β«1βγ (π‘)^((β1)/2) " " . sec^2β‘π₯ γ. ππ‘/sec^2β‘π₯ " " = β«1βγπ‘^((β1)/2) . ππ‘γ = π‘^(β 1/2 +1)/(β 1/2 +1) + πΆ = π‘^(1/2)/(1/2) + πΆ = γ2π‘γ^(1/2)+ πΆ = 2βπ‘+ πΆ = πβ(πππ§β‘π )+ πͺ (Using π‘=π‘ππβ‘π₯)