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Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=
2 Ιαν 2023 · 108. \quad \displaystyle ∫\frac {du} {u\sqrt {a+bu}}=\begin {cases} \frac {1} {\sqrt {a}}\ln \left|\frac {\sqrt {a+bu}−\sqrt {a}} {\sqrt {a+bu}+\sqrt {a}}\right|+C,\quad \text {if}\,a>0\\ [4pt] \frac {\sqrt {2}} {\sqrt {−a}}\tan^ {-1}\sqrt {\frac {a+bu} {−a}}+C,\quad \text {if}\,a<0 \end {cases}
Table of Basic Integrals. Basic Forms. 1 xndx = xn+1; n 6= 1. + 1. 1 dx. = ln jxj. Z. udv = uv. Z. vdu. (4) Z 1 1 dx = ln jax + bj ax + b a. Integrals of Rational Functions. (5) Z 1 1 dx = (x + a)2 x + a. (6) Z (x + a)n+1. (x + a)ndx = ; n 6= 1. n + 1. (7) Z (x + a)n+1((n + 1)x a) x(x + a)ndx = (n + 1)(n + 2) (8) 1 dx. + x2 = tan 1 x.
Integration by parts: u dv = uv − v du + C Partial Fractions: to integrate a function like ax+b (x+c)(x+d): Write ax+b (x+c)(x+d) = A (x+c) + B (x+d) = A(x+d)+B(x+c) (x+c)(x+d), so ax+b = A(x+d)+B(x+c)=(A+B)x+(Ad+Bc), so a = A+B and b = Ad+Bc; solve for A and B. The approach for more general denomenator can be found in nearly any calculus ...
A Review: The basic integration formulas summarise the forms of indefinite integrals for may of the functions we have studied so far, and the substitution method helps us use the table below to evaluate more complicated functions involving these basic ones.
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Integrals of Exponential and Logarithmic Functions. ∫ ln x dx = x ln x − x + C. + 1 x. + 1. x ∫ x ln xdx = ln x − + C. 2 + 1 ( n + 1 ) x dx = e x + C ∫.
