Example 6 (iii) - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
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Example 1 (i)
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Chapter 7 Class 12 Integrals
Serial order wise
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Example 6 Find the following integrals (iii) β«1β1/(1 + tanβ‘π₯ ) ππ₯ The given function cannot be integrated by direct substitution, so we have to simplify the given function . Simplifying the given function and integrating. β«1β1/(1 + tanβ‘π₯ ) .ππ₯ = β«1β1/(1 + sinβ‘π₯/cosβ‘π₯ ) .ππ₯ = β«1β1/((cosβ‘π₯ + sinβ‘π₯)/cosβ‘π₯ ) .ππ₯ (ππ πππ tanβ‘π₯=sinβ‘π₯/cosβ‘π₯ ) = β«1βcosβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ Multiplying and dividing by 2 = β«1β(2 cosβ‘π₯)/(2 (sinβ‘π₯ + cosβ‘π₯ ) ) .ππ₯ = β«1β(cosβ‘π₯ + cosβ‘π₯)/(2 (sinβ‘π₯ + cosβ‘π₯ ) ) .ππ₯ Adding and subtracting sinβ‘π₯ in the numerator = β«1β(cosβ‘π₯ + cosβ‘π₯ + sinβ‘π₯ β sinβ‘π₯)/(2 (sinβ‘π₯ + cosβ‘π₯ ) ) .ππ₯ = 1/2 β«1β(sinβ‘π₯ + cosβ‘π₯ + cosβ‘π₯ β sinβ‘π₯)/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ = 1/2 β«1β[(sinβ‘π₯ + cosβ‘π₯)/(sinβ‘π₯ + cosβ‘π₯ )+γcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ )] ππ₯ = 1/2 β«1β[1+γcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ )] ππ₯ = 1/2 [β«1βγ1.ππ₯+β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ )γ ππ₯] = 1/2 [π₯+πΆ1+β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) ππ₯] Take, I1 =β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ Let sinβ‘π₯ + cosβ‘π₯=π‘ Differentiate both sides π€.π.π‘.π₯. cosβ‘π₯βsinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) Putting value of (π ππβ‘π₯+πππ β‘π₯ ) and ππ₯ in I1 . I1=β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ I1 = β«1βγcosβ‘π₯ β sinγβ‘π₯/π‘ .ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) I1 =β«1β1/π‘ .ππ‘ Let sinβ‘π₯ + cosβ‘π₯=π‘ Differentiate both sides π€.π.π‘.π₯. cosβ‘π₯βsinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) Putting value of (π ππβ‘π₯+πππ β‘π₯ ) and ππ₯ in I1 . I1=β«1βγcosβ‘π₯ β sinγβ‘π₯/(sinβ‘π₯ + cosβ‘π₯ ) .ππ₯ I1 = β«1βγcosβ‘π₯ β sinγβ‘π₯/π‘ .ππ‘/(cosβ‘π₯ β sinβ‘π₯ ) I1 =β«1β1/π‘ .ππ‘ (ππ πππ β«1βγ1/π₯ .γ ππ₯=logβ‘γ |π₯|γ+πΆ) (ππ πππ π‘=sinβ‘π₯+cosβ‘π₯ ) (πβπππ πΆ=πΆ1/2+πΆ2/2)
