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Integration by parts
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Integration by parts
Last updated at May 29, 2023 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