Ex 7.2, 37 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Aug. 20, 2021 by Teachoo
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Ex 7.2, 37 You are here
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Transcript
Ex 7.2, 37 Integrate (π₯3 π ππ (tan^(β1)β‘γπ₯^4 γ))/(1 + π₯8) β«1βγ" " (π₯3 π ππ (tan^(β1)β‘γπ₯^4 γ ))/(1 + π₯8)γ . ππ₯ Let tan^(β1)β‘γπ₯^4 γ= π‘ Differentiating both sides π€.π.π‘.π₯ 1/(1 +(π₯^4 )^2 ). π(π₯^4 )/ππ₯= ππ‘/ππ₯ 1/(1 +π₯^8 ). 4π₯^3=ππ‘/ππ₯ (4π₯^3)/(1 + π₯^8 )=ππ‘/ππ₯ (Using (π(γπ‘ππγ^(β1)β‘π₯))/ππ₯=1/(1 + π₯^2 ) and chain rule ) ππ₯=(1 + π₯^8)/(4π₯^3 ) . ππ‘ Now, our function becomes β«1βγ" " (π₯3 π ππ (tan^(β1)β‘γπ₯^4 γ ))/(1 + π₯8)γ . ππ₯ Putting γπ‘ππγ^(β1)β‘γπ₯^4 γ=π‘ & ππ₯=(1 + π₯^8)/(4π₯^3 ) . ππ‘ = β«1βγ" " (π₯^3 π ππ (π‘))/(1 + π₯^8 )γ. (1 + π₯^8)/(4π₯^3 ) ππ‘" " = β«1βγ" " sinβ‘π‘/4γ ππ‘" " = 1/4 β«1βsinβ‘π‘ . ππ‘" " = (β1)/4 cosβ‘π‘+ πΆ = (βπ)/π γπππ γβ‘(γπππ§γ^(βπ)β‘γπ^π γ )+πͺ (Using π‘=γπ‘ππγ^(β1)β‘γπ₯^4 γ)
