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Chapter 7 Class 12 Integrals

Serial order wise

Last updated at May 29, 2023 by Teachoo

Misc 24 Evaluate the definite integral β«_(π/2)^πβγe^π₯β‘((1 βsinβ‘π₯)/(1 βcosβ‘π₯ )) ππ₯γ β«_(π/2)^πβγe^π₯β‘((1 β sinβ‘π₯)/(1 βγ cosγβ‘π₯ )) ππ₯γ = β«_(π/2)^πβγe^π₯β‘((1 )/(1 β cosβ‘π₯ )βsinβ‘π₯/(1 β cosβ‘π₯ )) ππ₯γ = β«_(π/2)^πβγe^π₯β‘((βsinβ‘π₯)/(1 β cosβ‘π₯ )+(1 )/(1 β cosβ‘π₯ )) ππ₯γ Let f(x) = (βsinβ‘π₯)/(1 β cosβ‘π₯ ) fβ(x) = β[(cosβ‘π₯ (1 β cosβ‘γπ₯) β (sinβ‘γπ₯) (sinβ‘γπ₯)γ γ γ)/((1 β γcosβ‘π₯)γ^2 )] =β[(cosβ‘γπ₯ βγ γπππ γ^2 π₯ β γπ ππγ^2 π₯)/((1 β γcos π₯β‘)γ^2 )] = β(cosβ‘γπ₯ β (γπππ γ^2 π₯ + γπ ππγ^2 π₯)γ/((1 β γπππ π₯β‘)γ^2 )) = β (cosβ‘γπ₯ β 1γ/γ(1 βγ cosγβ‘γπ₯)γγ^2 ) = ( 1 β cosβ‘γπ₯ γ)/γ(1 β cosβ‘γπ₯)γγ^2 = 1/(1 β cosβ‘π₯ ) Hence, the given integration is of form, β«1βγπ^π₯ (π(π₯)+π^β² (π₯)) ππ₯γ= π^π₯ π(π₯) Where f(x) = (βsinβ‘π₯)/(1 β cosβ‘π₯ ) and fβ(x) = 1/(1 β cosβ‘π₯ ) Hence, β«_(π/2)^πβγe^π₯β‘((1 β sinβ‘π₯)/(1 βγ cosγβ‘π₯ )) ππ₯γ = β«_(π/2)^πβγe^π₯β‘((1 )/(1 β cosβ‘π₯ )βsinβ‘π₯/(1 β cosβ‘π₯ )) ππ₯γ = [π^π₯ ((βsinβ‘π₯)/(1 βcosβ‘π₯ ))]_(π/2)^π = ((π^π₯ sinβ‘π₯)/cosβ‘γπ₯ β1γ )_(π/2)^π Putting limits = (π^π sinβ‘π)/cosβ‘γπ β1γ β (π^(π/2) sinβ‘γπ/2γ)/cosβ‘γ π/2 β1γ = (π^π Γ (0))/(β1 β1)β (π^(π/2) (1))/(0 β 1) = π^(π /π)