Definite Integration by properties - P4

Chapter 7 Class 12 Integrals
Concept wise

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Example 30 Evaluate β«_0^πβ(π₯ π ππ π₯)/(1 + cos^2β‘π₯ ) ππ₯ Let I=β«_0^πβ(π₯ sinβ‘π₯)/(1 + cos^2β‘π₯ ) ππ₯ β΄ I=β«_0^πβ((π β π₯) π ππ(π β π₯))/(1 + cos^2β‘(π β π₯) ) ππ₯ I=β«_0^πβγ((π β π₯) sinβ‘π₯)/(1 + [βcosβ‘π₯ ]^2 ) ππ₯γ I=β«_0^πβγ(π sinβ‘γπ₯ β π₯ sinβ‘π₯ γ)/(1 + cos^2β‘π₯ ) ππ₯γ Adding (1) and (2) i.e. (1) + (2) I +I=β«_0^πβ(π₯ sinβ‘π₯)/(1 + cos^2β‘π₯ ) ππ₯+β«_0^πβ(π sinβ‘γπ₯ β π₯ π πππ₯γ)/(1 + cos^2β‘π₯ ) ππ₯ 2I =β«_0^πβ(π₯ sinβ‘γπ₯ + π sinβ‘γπ₯ β π₯ sinβ‘π₯ γ γ)/(1 + cos^2β‘π₯ ) ππ₯ 2I =β«_0^πβ(π sinβ‘γπ₯ γ)/(1 + cos^2β‘π₯ ) ππ₯ 2I =πβ«_0^πβsinβ‘γπ₯ γ/(1 + cos^2β‘π₯ ) ππ₯ β΄ I = π/2 β«_0^πβγsinβ‘π₯/(1 + cos^2β‘π₯ ) ππ₯γ Let cos π₯ = t Differentiate both sides w.r.t.π₯ β sinβ‘γπ₯=ππ‘/ππ₯γ ππ₯=ππ‘/(βsinβ‘π₯ ) Putting the values of (cosβ‘π₯ ) and dπ₯, we get I =π/2 β«1_0^πβγsinβ‘π₯/(1 + π‘^2 ).ππ₯γ I = π/2 β«1_0^πβγsinβ‘π₯/(1 + π‘^2 ) Γ ππ‘/(βsinβ‘π₯ )γ I = (βπ)/2 β«1_0^πβγ1/(1 + π‘^2 ) .ππ‘γ I = (βπ)/2 β«1_0^πβγ1/((1)^2 + (π‘)^2 ) .ππ‘γ I=(βπ)/2 [1/1 tan^(β1)β‘(π‘/1) ]_π^π =(βπ)/2 [tan^(β1)β‘(π‘) ]_0^π Putting t = cos x =(βπ)/2 [tan^(β1)β‘(cosβ‘π₯ ) ]_0^π =(βπ)/2 [tan^(β1)β‘γ[cosβ‘π ]βtan^(β1)β‘[cosβ‘0 ] γ ] =(βπ)/2 [tan^(β1)β‘γ(β1)βtan^(β1)β‘(1) γ ] =(βπ)/2 [βtan^(β1)β‘γ(1)βtan^(β1)β‘(1) γ ] =(βπ)/2 [β2 tan^(β1)β‘(1) ] =π[tan^(β1)β‘(1) ] = π[π/4] =π^π/π