

Examples
Last updated at Dec. 16, 2024 by Teachoo
Transcript
Example 21 Find β«1βπ^π₯ sinβ‘π₯ ππ₯ Let I1 = β«1βγ π^π₯ γ sinβ‘π₯ ππ₯ I1 = sinβ‘π₯ β«1βγπ^π₯ ππ₯γββ«1β(π(sinβ‘π₯ )/ππ₯ β«1βγπ^π₯ ππ₯γ) ππ₯ I1 = π^π₯ sinβ‘π₯ββ«1βγcosβ‘π₯ . π^π₯ ππ₯γ 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) = sin x and g(x) = ex Solving I2 I2 = β«1βγcosβ‘π₯ . π^π₯ ππ₯γ I2 = cos x β«1βγπ^π₯ ππ₯γ β β«1βγ((cosβ‘π₯)β²γ β«1βγπ^π₯ ππ₯γ)ππ₯ I2 = cos x π^π₯ β β«1βγ(βsinβ‘π₯)γ π^π₯ ππ₯ I2 = π^π₯ cos x + β«1βsinβ‘π₯ π^π₯ ππ₯ I2 = π^π₯ cos x + πΌ1 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) = sin x and g(x) = ex Now, Putting value of I2 in (1) , I1 = " " π^π₯ sinβ‘π₯ββ«1βγcosβ‘π₯ π^π₯ γ ππ₯ I1 = " " π^π₯ sinβ‘π₯β(π^π₯ cosβ‘π₯+πΌ1)+πΆ I1 = " " π^π₯ sinβ‘π₯βπ^π₯ cosβ‘π₯βπΌ1+πΆ 2I1 = " " π^π₯ sinβ‘π₯βπ^π₯ cosβ‘π₯ + πΆ I1 = 1/2 (π^π₯ sinβ‘π₯βπ^π₯ cosβ‘π₯ ) + C π°π = π^π/π (πππβ‘πβπππβ‘π ) + C