University:
California State University, Northridge
Course:
MATH 150B | Calculus II
Academic year:

2023
Views:

372

Pages:

4
Author:

nadiascoot22

Gx4 (x² In Sx) = (x'] In(5x)+ [In sx] x² 3x In (5x)+ 5x 3x In (Sx) + 5x = = = 3x In (Sx) + x¹ P Ex5 (In cosx)] 1 COS X Cos x Simx COSX Sinx = [cos(x)] = : tan x 6 x(2x+1) = [Inx² + In (2x²+1) - (nds-x) = [2 Inx + 3 ln (2x²+1) - In(5-x)] =2*+34 Ex 1 y= (x-1) 2 In (x-1) In y = In (x-1) - ((2x-1) Iny = 4 In (x-1) - In (2x-1) dx =[] 2 3(2x-1) Thp 2 3(2x-1) 32x-1 # [Iny] = {4 In(x-1) - 1n(2x-1)) [y]=4(1)-(2) 4 x-1 3 2x Ex 2 In- siny + x²=0 ax [In 2x-Iny - simy +x)= [0] dx - 2- y' - cosy.y' + 2x 0 - *y cosy y + 2x=0 y'(cosy)=-2x- y'= -cosy Intergral: Südu = mlu+c 4= 5x-2 du Sdx dy= dx Gx1 Ssx dx 5x-2 = m15x-21+ c S→ Südumlul+c Ex2dxadu Sudu 4. Inx dudx x du = dx " 3/ = (mx) + C S Sin x dx COS X + C Stan u du In / secu/+c S cos x de = sinx +c Scot u du = In smul+c S secx dx tan x te S sec u du = In / secuttmy+c S sec tan x dx = secx + C Scse u du = In/cscu - cotul+C Jcscx cotx dx=cscx+c Scscxdx=-cotx +c 1 Ex S S sec'+ Secx tanp Secx + tanx Ex Sx sec (x') dx = Sx sec u = S sec u du In/secu + tanal to In | Secx' + tanx/+c Secx dx du = 2xdx 1 1 1 dr da: (secx tanx + seca) dx = 2 In+c = S secx Secx + tans Secx 1 tams dx S secx dx In secx + tanx/+C Ex S Sin 2x 1+ sin²x dx 4= 1 + sinx du2sinxcosx dx Sa sinx cosx I+ sinx dx Ex S tanx dx = S sinx dx =-S du =-In/u/+c =-In cosx + c = In leeset" to du-sinx dx du = dx x307 cosx secx S du In lul +C = In I+ sinx +c = n 6.1 Natural Log Function Review log.x=4-2-x expand Ex 1. 3x 29 1. (n1=0 2. In ay 3. (n= 4. In x = Inx + Iny Lux - Iny r lnx In 3x - In 29 = ln3+ Lux - In2 - Iny = 173 + 4 lnx - 1n2-2 Iny Ex2. In x+1 = In (x²+1)- In x In(x²+1) - Inx Graph f(x) = In x g(a)= ex 3 are they inverse Ex 3. In x Sin(x) 3 ((4) [x] = 5x In x sin(x) Inx+2 In x³ + In sin(x) - In (x+2) = 5 in x + (n sin (πι)-In (x+2) = 5 (nx + 2 (n sin (x) - In (x+2) = Ex 5 ∫ x dx = + c 4 Exb S 4x dx = 45xdx = 4+ = x²+c Ex4. [5 lnx + In (x²+1)-4(n (x-2)] == [Inx² + In(x²+1)²-(n(x-2)") 5 = [In (x-2) (x-2)