Example 20 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 20 Find β«1β(π₯ sin^(β1)β‘π₯)/β(1 β π₯^2 ) ππ₯ Example 20 Find β«1β(π₯ sin^(β1)β‘π₯)/β(1 β π₯^2 ) ππ₯ β«1β(π₯ sin^(β1)β‘π₯)/β(1 β π₯^2 ) ππ₯ Let t = γπ ππγ^(β1) (π₯) ππ‘/ππ₯=1/β(1 β π₯^2 ) dt = ππ₯/β(1 β π₯^2 ) So, our equation becomes β«1β(π₯ sin^(β1)β‘π₯)/β(1 β π₯^2 ) ππ₯ = β«1βγsinβ‘γπ‘Γπ‘ γΓππ₯/β(1 β π₯^2 )γ = β«1βγsinβ‘γπ‘Γπ‘ γ ππ‘γ =π‘ β«1βγsinβ‘γπ‘ ππ‘ β β«1β(π(π‘))/ππ‘γ γ β«1βsinβ‘γπ‘ ππ‘ γ ππ‘ = t (βcos t) β β«1β(βcosβ‘π‘ ) ππ‘ = β t cos t + β«1βcosβ‘π‘ ππ‘ = βt cos t + sin t + C Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = t and g(x) = sin t Hence we take First function :-f(x) = t Second function :- g(x) = sin t β«1βγsinβ‘γπ‘Γπ‘ ππ‘=π‘ β«1βγsinβ‘γπ‘ ππ‘ β β«1β(π(π‘))/ππ‘γ γγ γ β«1βsinβ‘γπ‘ ππ‘ γ ππ‘ = t (βcost) β β«1β(βcosβ‘π‘ ) ππ‘ = β t cost + β«1βcosβ‘π‘ ππ‘ = βt cost + sin t + C (β«1βsinβ‘γπ₯ ππ₯=βcosβ‘π₯ γ " " ) (β«1βcosβ‘γπ₯ ππ₯ = sinβ‘π₯ γ " " ) t = γπ ππγ^(β1) (π₯) sin t = x sin t = x γπ ππγ^2 π‘ = π₯^2 1 β cos^2β‘π‘ = π₯^2 γπππ γ^2 t = 1 β π₯^2 cos t = β(1βπ₯^2 ) Now, Hence putting the values. β«1β(π₯ γπ ππγ^(β1) π₯)/β(1βπ₯^2 ) ππ₯=" βt cost + sin t + C" =πββ(πβπ^π ) γπππγ^(βπ) π +π
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo