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Will try integration by parts.
uv - int[v du]
u = sec(x)----------------du = sec(x) tan(x)
dv = tan(x)---------------v = ln[sec(x)]
sec(x) ln[sex(x)] - int[lnsec(x) dx]
= sec(x) ln[sec(x)] - xlnsec(x) - x + C
===========================
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∙ 12y ago
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Integration by parts of x 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 -----------------------------------------
How do you simplify cosx plus sinx tanx?
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
What is the derivative of secxtanx?
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)
What is the derivative of 5secx?
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
Integration of root tanx?
The integral of sqrt(tan(x)) is rather complex and is hard to show with the formatting allowed on Answers.com. See the related links for a representation of the answer.
What is the integral of tan cubed x secx dx?
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
Intergrate sec x?
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
Integration by parts of x 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 -----------------------------------------
How do you Prove sin x times sec x equals tan x?
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
What is the integral of tanx times sqrt secx dx?
See related link below for answer
How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(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
How do you simplify cosx plus sinx tanx?
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
What is the derivative of secxtanx?
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)
What is secx?
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.
What is the derivative of 5secx?
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
