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MATH 142 Final Exam: Integration Techniques and Trigonometric Substitutions, Exams of Mathematics

Solutions for various integration problems using trigonometric substitutions in the context of a final exam for math 142. Topics covered include integrals of the form (tan x sec x dx), (sin x cos x dx), and (sin(α ± β) dx), as well as techniques for evaluating definite integrals using upper and lower limits. The document also includes information on finding antiderivatives using substitution and the evaluation of definite integrals using limits.

Typology: Exams

Pre 2010

Uploaded on 10/01/2009

koofers-user-sox-1
koofers-user-sox-1 🇺🇸

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Download MATH 142 Final Exam: Integration Techniques and Trigonometric Substitutions and more Exams Mathematics in PDF only on Docsity! MATH 142 – Final Exam √ a2 − x2 → x = a sin t − π 2 ≤ t ≤ π 2 → a2−x2 = a2−a2 sin2 t = a2 cos2 t √ a2 + x2 → x = a tan t − π 2 < t < π 2 → a2+x2 = a2+a2 tan2 t = a2 sec2 t √ x2 − a2 → x = a sec t { 0 ≤ t < π/2 if x ≥ a π/2 < t ≤ π if x ≤ a → x 2− a2 = a2 sec2 t− a2 = a2 tan2 t ∫ tanm x secn x dx ⇒    n even → u = tanx → sec2 x = tan2 x+ 1 m odd → u = sec x → tan2 x = sec2 x− 1 m even and n odd → tan2 x = sec2 x− 1 ∫ sinm x cosn x dx ⇒    n odd → u = sin x → cos2 x = 1− sin2 x m odd → u = cos x → sin2 x = 1− cos2 x m even and n even → { sin2 x = 1 2 (1− cos 2x) cos2 x = 1 2 (1 + cos 2x) sinα cos β = 1 2 [sin(α− β) + sin(α + β)] sinα sin β = 1 2 [cos(α− β)− cos(α + β)] cosα cos β = 1 2 [cos(α− β) + cos(α + β)] sin(α + β) = sinα cos β + cosα sin β sin(α− β) = sinα cos β − cosα sin β cos(α + β) = cosα cos β + sinα sin β cos(α− β) = cosα cos β − sinα sin β sin(2θ) = 2 sin θ cos θ sin2 θ + cos2 θ = 1 ∫ sinn x dx = − 1 n sinn−1 x cosx+ n− 1 n ∫ sinn−2 x dx ∫ cosn x dx = 1 n cosn−1 x sinx+ n− 1 n ∫ cosn−2 x dx
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