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Ex 7.9, 20 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Ex 7.9, 20 β«_0^1β(π₯ π^π₯+sinβ‘γππ₯/4γ ) ππ₯ Let F(π₯)=β«1β(π₯π^π₯+π ππ ππ₯/4)ππ₯ =β«1βγπ₯π^π₯ ππ₯+β«1βγsinβ‘(ππ₯/4) ππ₯γγ Solving I1 and I2 separately Solving π°π β«1βγπ₯π^π₯ ππ₯γ =π₯β«1βγπ^π₯ ππ₯ββ«1β[(ππ₯/ππ₯) β«1βγπ^π₯ ππ₯γ]ππ₯γ =π₯π^π₯ββ«1β(1.π^π₯ ππ₯)ππ₯ =π₯π^π₯ββ«1βγπ^π₯ ππ₯γ =π₯π^π₯βπ^π₯ =π^π₯ (π₯β1) Solving by parts β«1βγπ(π₯)π(π₯)ππ₯=π(π₯)β«1βγπ(π₯)ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯)ππ₯)ππ₯γγ Let π(π₯)=π₯ and π(π₯)=π^π₯ Solving I2 β«1βγsinβ‘(ππ₯/4) ππ₯γ = 1/(π/4) (βcosβ‘(ππ₯/4) ) = (β4)/π cosβ‘(ππ₯/4) Therefore, F(π₯)=β«1βγπ₯π^π₯ ππ₯+β«1βγπ ππ π/4 π₯ ππ₯γγ =π^π₯ (π₯β1)β4/π cosβ‘(ππ₯/4) Now, β«_0^1β(π₯π^π₯+π ππ ππ₯/4) ππ₯=πΉ(1)βπΉ(0) =(π^1 (1β1)β4/π cosβ‘((π Γ 1)/4) )β(π^0 (0β1)+4/π πππ ((π Γ 0)/4)) =πΓ0β4/π πππ π/4β1(β1)+4/π cosβ‘0 =(β4)/( π) πππ π/4+1+4/π =(β4)/( π) 1/β2+1+4/π =(β2β2)/( π) +1+4/π =π+π/π β(πβπ)/π
