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Q: What is the integral of tan squared x?

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tan(sqrtX) + C

Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C

Yes. Both expressions are the same.

âˆ« 1/cos2(x) dx = tan(x) + C C is the constant of integration.

-cotan(x)

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

integral of radical sinx

22

ln |sec x + tan x| + C

ln|sec x + tan x| + C.

The indefinite integral of (1/x^2)*dx is -1/x+C.

maths signs

arctan(x)

tan(x) + C d/dx tan(x) = d/dx (sin(x))/(cos(x)) = (sin^2(x)+cos^2(x))/(cos^2(x)) = 1/(cos^2(x)) = sec^2(x) NEVER FORGET THE CONSTANT!

.5(x-sin(x)cos(x))+c

the integral of the square-root of (x-1)2 = x2/2 - x + C

There can be no definite integral because the limits of integration are not specified. The indefinite integral of 1/x2 is -1/x + C

cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.

Given y = tan x: dy/dx = sec^2 x(secant of x squared)

The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1

(x+sinxcosx)/2, can do it by parts or by knowing your double angle formulas

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

ln(sinx) + 1/3ln(sin3x) + C

- ln ((x^2)-4)

âˆ« tan(x) dx = -ln(cos(x)) + C C is the constant of integration.