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 ===========================
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
It is minus 1 I did this: sinx/cos x = tan x sinx x = cosx tanx you have (x - sinxcosx) / (tanx -x) (x- cos^2 x tan x)/(tanx -x) let x =0 -cos^2 x (tanx) /tanx = -cos^x -cos^2 (0) = -1
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.
One can use integration by parts to solve this. The answer is (x-1)e^x.
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 ===========================
(-x+tanx)'=-1+(1/cos2x)
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
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
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
It is minus 1 I did this: sinx/cos x = tan x sinx x = cosx tanx you have (x - sinxcosx) / (tanx -x) (x- cos^2 x tan x)/(tanx -x) let x =0 -cos^2 x (tanx) /tanx = -cos^x -cos^2 (0) = -1
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
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.
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
Integration by Parts is a special method of integration that is often useful when two functions.
sec^2(x)