Sep 27, 2015 · d/dx2^(sin(pix))=2^(sin(pix))*ln2*cospix*(pi) Using the following standard rules of differentiation: d/dxa^(u(x))=a^u*lna*(du)/dx d/dx sinu(x) = cosu(x)*(du)/dx d ...

Oct 16, 2015 · Add the form int sin^m u cos^n u du with at least one of m, n odd to your mathematical cookbook. int sin^m u cos^n u du with at least one of m, n odd. Integrate by …

Aug 8, 2016 · 1 lim_{x to 1} sin((pix)/2) =sin((pi)/2) Well, sin((pi)/2) = 1 and the function f(t) = sin t is also continuous in t in (oo, oo)

Feb 26, 2017 · Trapezium rule gives: int_2^5 \ sin(pix) \ dx ~~ 0.629062 The values of f(x)=sin(pix) are tabulated as follows (using Excel) working to 5dp Using the trapezoidal rule ...

Apr 30, 2015 · In this way: y'=pi*(cospix*pi)=pi^2cospix y''=pi^2(-sinpix*pi)=-pi^3sinpix.

Use: #x^3-8 = (x-2)(x^2+2x+4)# and . #sin(pix) = sin(pix-2pi) = sin(pi(x-2))# #(x^3-8)/sin(pix) = ((x-2)(x^2+2x+4))/sin(pi(x-2))#

Jan 10, 2018 · 3217 views around the world You can reuse this answer ...

Jan 9, 2018 · lim_(x->1)sin(pix)/sin(3pix)=1/3 lim_(x->1)sin(pix)/sin(3pix) This limit is in the form 0/0, so we can use l'Hôpital's rule, which states that, if a limit lim_(x->c ...

See the explanation: First, find the amplitude and the period and phase shifting: a sin bx+c amplitude: |a| period: for sine its period is 2pi so (2pi)/b phase shift: -c So amplitude = |-2|=2 …

Jul 28, 2015 · So here we go: For g(x) = -2sin(pix+pi/3), We have Amplitude = 2" " (The minus sign will flip the graph across the x axis.) Period is (2pi)/pi = 2 Phase Shift is the solution to …