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What is the integral of tan squared x?
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
What is 1 squared radical 40 plus 3 squared radical 10 equal?
1 squared radical 40 is equal to 1 squared radical (4 x 10), and this become 2 squared radical 10. By adding 2 squared radical10 with 3 squared radical 10 we get 5 squared radical 10.
What is the integral of sec squared 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!
What is integral of SECx?
ln |sec x + tan x| + C
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
What is the integral sec x?
ln|sec x + tan x| + C.
What is the integral of the sine of x divided by the cosine squared of x with respect to x?
∫ sin(x)/cos2(x) dx = sec(x) + C C is the constant of integration.
What is the integral of tan squared x?
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
What is 1 squared radical 40 plus 3 squared radical 10 equal?
1 squared radical 40 is equal to 1 squared radical (4 x 10), and this become 2 squared radical 10. By adding 2 squared radical10 with 3 squared radical 10 we get 5 squared radical 10.
What is the integral of cosec squared x?
-cotan(x)
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 integral of sec squared 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!
What is integral of SECx?
ln |sec x + tan x| + C
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
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
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
Integral of sec2x-cosx plus x2dx?
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
What is the integral of 1 over x squared plus 1?
arctan(x)
