integral of radical sinx
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!
ln |sec x + tan x| + C
The indefinite integral of (1/x^2)*dx 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.
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
tan(sqrtX) + C
∫ 1/cos2(x) dx = tan(x) + C C is the constant of integration.
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
Yes. Both expressions are the same.
-cotan(x)
integral of radical sinx
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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!
ln|sec x + tan x| + C.
ln |sec x + tan x| + C
The indefinite integral of (1/x^2)*dx is -1/x+C.