Αποτελέσματα Αναζήτησης
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=
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 ∫.
Table of Basic Integrals Basic Forms (1) Z xndx= 1 n+ 1 xn+1; n6= 1 (2) Z 1 x dx= lnjxj (3) Z udv= uv Z vdu (4) Z 1 ax+ b dx= 1 a lnjax+ bj Integrals of Rational Functions (5) Z 1 (x+ a)2 dx= 1 x+ a (6) Z ... Integrals with Trigonometric Functions (71) Z sinaxdx= 1 a cosax (72) Z sin2 axdx= x 2 sin2ax 4a (73) Z sin3 axdx= 3cosax 4a + cos3ax 12a ...
Table of Basic Integrals1 (1) Z xn dx = 1 n+1 xn+1; n 6= 1 (2) Z 1 x dx = lnjxj (3) Z u dv = uv Z vdu (4) Z e xdx = e (5) Z ax dx = 1 lna ax (6) Z lnxdx = xlnx x (7) Z sinxdx = cosx (8) Z cosxdx = sinx (9) Z tanxdx = lnjsecxj (10) Z secxdx = lnjsecx+tanxj (11) Z sec2 xdx = tanx (12) Z secxtanxdx = secx (13) Z a a2 +x2 dx = tan 1 x a (14) Z a a2 ...
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 ...
Table of Integrals 1. Z ur du = ur+1 r +1 +C, r 6=−1 2. Z 1 u du =ln|u|+C 3. Z eu du = eu +C 4. Z sinudu= − cosu+C 5. Z cosudu= sinu+C 6. Z tanudu=ln|secu|+C 7. Z cotudu=ln|sinu|+C 8. Z secudu=ln|secu+tanu|+C 9. Z cscudu=ln|cscu−cotu|+C 10. Z secutanudu= secu+C 11. Z cscucotudu= − cscu+C 12. Z sec2 udu= tanu+C 13. Z csc2 udu= − cotu+C ...
Table of Integrals 1. Z [u(x)]ru0(x)dx = ur+1(x) r +1 +C, r 6=−1 2. Z u0(x) u(x) dx =ln|u(x)|+C 3. Z eu(x)u0(x)dx = eu(x) +C 4. Z sin[u(x)]u0(x)dx = −cos u(x)+C 5. Z cos[u(x)]u0(x)dx = sin u(x)+C 6. Z tan[u(x)]u0(x)dx =ln|sec u(x)|+C 7. Z cot[u(x)]u0(x)dx =ln|sin u(x)|+C 8. Z sec[u(x)]u0(x)dx =ln|sec u(x)+tanu(x)|+C 9. Z csc[u(x)]u0(x)dx ...
