Integration by specific formulaes - Formula 5

Chapter 7 Class 12 Integrals
Concept wise

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### Transcript

Example 40 Evaluate β«1β(sinβ‘γ2π₯ cosβ‘2π₯ γ ππ₯)/(β(9 βcos^4β‘(2π₯) ) ) ππ₯ β«1β(sinβ‘γ2π₯ cosβ‘2π₯ γ ππ₯)/(β(9 β cos^4β‘2π₯ ) ) ππ₯ Let cos^2β‘2π₯=π‘ Differentiating both sides w.r.t.π₯ β2.2 cosβ‘2π₯ sinβ‘2π₯=ππ‘/ππ₯ ππ₯=ππ‘/(β4 cosβ‘2π₯ sinβ‘2π₯ ) Hence, our equation becomes β«1β(sinβ‘γ2π₯ cosβ‘2π₯ γ )/(β(9 β cos^4β‘2π₯ ) ) ππ₯ =β«1βγsinβ‘γ2π₯ cosβ‘2π₯ γ/β(9 β π‘^2 ) ππ₯γ =β«1βγ(sinβ‘2π₯ cosβ‘2π₯)/β(9 β π‘^2 ) Γ ππ‘/(β4 cosβ‘2π₯ sinβ‘2π₯ )γ =1/(β4) β«1βππ‘/β((3)^2 β (π‘)^2 ) =(β1)/( 4) [sin^(β1)β‘γπ‘/3+πΆ1γ ] =(β1)/( 4) π ππ^(β1) π‘/3βπΆ1/4 =(βπ)/( π) πππ^(βπ) [π/π πππ^π ππ]+πͺ