Integration using trigo identities - sin^2,cos^2 etc formulae

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.3, 24 β«1β(π^π₯ (1 + π₯))/(cos^2β‘(π^π₯ π₯) ) ππ₯ equals (A) βcotβ‘(ππ₯^π₯ ) + πΆ (B) tanβ‘(π₯π^π₯ ) + πΆ (C) tanβ‘(π^π₯) + πΆ (D) cotβ‘(π^π₯) + πΆ β«1β(π^π₯ (1 + π₯))/cos^2β‘(π₯π^π₯ ) ππ₯ Put γπ₯πγ^π₯=π‘ Differentiating w.r.t.x π(π₯)/ππ₯ . π^π₯+π(π^π₯ )/ππ₯ . π₯=ππ‘/ππ₯ π^π₯+(π^π₯ ). π₯=ππ‘/ππ₯ Using product rule as (π’π£)^β²=π’^β² π£+ π£^β² π’ π^π₯ (1+π₯)=ππ‘/ππ₯ ππ₯=ππ‘/(π^π₯ (1 + π₯) ) Thus, our equation becomes β«1β(π^π₯ (1 + π₯))/cos^2β‘(π₯π^π₯ ) ππ₯ = β«1β(π^π₯ (1 + π₯))/cos^2β‘(π‘) Γ ππ‘/(π^π₯ (1 + π₯) ) =β«1βππ‘/cos^2β‘π‘ . ππ‘ =β«1βsec^2β‘π‘ . ππ‘ =tanβ‘π‘+πΆ Putting value of π‘=π₯π^π₯ =tanβ‘(π₯π^π₯ )+πΆ β΄ B is correct answer ( As β«1βsec^2β‘π₯ ππ₯=tanβ‘π₯+πΆ)