It isn't clear what you mean with "by". Are you multiplying 1 by secx, or perhaps dividing? Also, is the multiplication (or division) only by sec x, or by the sum of secx + cos x?
-(10/x)
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
It isn't clear what you mean with "by". Are you multiplying 1 by secx, or perhaps dividing? Also, is the multiplication (or division) only by sec x, or by the sum of secx + cos x?
This is a trigonometric integration using trig identities. S tanX^3 secX dX S tanX^2 secX tanX dX S (secX^2 -1) secX tanX dX u = secX du = secX tanX S ( u^2 - 1) du 1/3secX^3 - secX + C
ln |sec x + tan x| + C
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
-5
secx is the inverse of cosx. secx=1/cosx. A secant is also a line drawn through the graph that touches two points on a function.
d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)
e^x/1-e^x
-(10/x)
Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C