∫ tan(x) dx =
∫ sin(x)/cos(x) dx
Let cos(x) = u
Therefore, sin(x) = -du/dx >>>> (Derivative of cos(x))
∫ (-du/dx)/u dx >>>>>>>>>>> Substitute the values
∫ -du/udx dx
∫ -du/u >>>>>>>>>>>>>>>>> dx * 1/dx cancel each other out
∫ -1/u du
= -ln(u) + C
= -ln(cos(x)) + C >>>>>>>>>>> Substitute cos(x) back into u
See related link below for answer
It isn't clear what you mean by "underroot"; I am not aware of a mathematical function by that name.
Tanx was created in 1972-10.
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
(-x+tanx)'=-1+(1/cos2x)
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
See related link below for answer
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
It isn't clear what you mean by "underroot"; I am not aware of a mathematical function by that name.
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.
Tanx was created in 1972-10.
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
(-x+tanx)'=-1+(1/cos2x)
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1
d/dx(1+tanx)=0+sec2x=sec2x