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It isn't clear what you mean by "underroot"; I am not aware of a mathematical function by that name.

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โˆ™ 2017-04-25 17:07:49
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Q: What is the integral of underroot tanx plus underroot cotxdx?
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Integration of tan pow4x?

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

What is the Derivative of -x plus tanx?


What is the integral of tanx times sqrt secx dx?

See related link below for answer

What is the derivative of 1 plus tanx?


How do you simplify tanx plus 1 sqaured?

I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1

Tan plus cot divided by tan equals csc squared?

(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.

Tan3x plus tanx equals 2tan2x?


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)

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

What is the differentiate of y equals tanx plus c?

y' = (sec(x))^2

When was Tanx created?

Tanx was created in 1972-10.

How do you simplify tanx plus tany per cosx plus coty?

(tan x- 1)/ (1+tan x)

Integration of root tanx?

The integral of sqrt(tan(x)) is rather complex and is hard to show with the formatting allowed on See the related links for a representation of the answer.

Intergrate sec x?

Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|

Integration of square root tan x dx?

for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..

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...:))

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

Does sin x tan x plus cos x equals tan x?

NO, sinxtanx=sinxsinx/cosx since tanx is sinx/cosx this is sin^2xcosx now add cosx cosx(sin^2x+1) after factoring Does this equal tanx? No, since this would require tanx to equal cosx(sin^2x+1) and it does not.

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

How do you prove the following identity sec x - cos x equals sin x tan x?

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

Tanx equals 45 is 49 degrees?


How do you differentiate Tanx?


Which is equivalent to (1- sin2x)tanx?


How do you integrate sec cube 2x?

The integral of sec(x) is ln|secx+tanx| + C Since the derivative is taken to the third power, we have to consider the chain rule; the original equation must be to the fourth power, and in order for that to be canceled out, the equation must also have had a coefficient of 1/4. 2x is also subject to the chain rule. I would suggest u substitution. integral(sec(2x))^3 dx u=2x du=2dx dx=1/2du integral (sec(u))^3 *1/2 du 1/8 secxtanx + 1/8(ln|secx+tanx|^4) + C

What is the limit of x- sin x cos x over tan x -x as x tends to zero?

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