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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
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What is the volume of a cube with sides of 2x?
2x x 2x x 2x = 8x3
What do you do to find the surface area of a shape?
Rectangular Prisim 2x(area of front)+2x(area of side)+x2(area of length) Cube 2x(area of side)+2x(area of side)+2x(area of side)
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
How do you differentiate cos squared x?
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
How do you integrate of e -2x?
To integrate e^(-2x)dx, you need to take a u substitution. u=-2x du=-2dx Since the original integral does not have a -2 in it, you need to divide to get the dx alone. -(1/2)du=dx Since the integral of e^x is still e^x, you get: y = -(1/2)e^(-2x) Well, that was one method. I usually solve easier functions like this by thinking how the function looked like before it was differentiated. I let f(x) stand for the given function and F(x) stand for the primitive function; the function we had before differentiation (the integrated function). f(x)= e-2x <-- our given function F(x)= e-2x/-2 <-- our integrated function Evidence: F'(x)= -2e-2x/-2 = e-2x = f(x) Q.E.D It's as simple as that.
How do you integrate e 2x?
e 2x = (1/2) e 2x + C ============
How do you integrate e powerintegral2x-1?
2x
How do you integrate functions?
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
How do you integrate of e power 2x plus e power minus 2x?
int[e(2X) +e(- 2X)] integrate term by term 1/22 e(2X) - 1/22 e(- 2X) + C (1/4)e(2X) - (1/4)e(- 2X) + C ====================
What is the derivative of secant 2X?
d/dx sec(2x) = 2sec(2x)tan(2x)
Which functions have a period of Pi?
sin(2x), cos(2x), cosec(2x), sec(2x), tan(x) and cot(x) are all possible functions.
What is sec2x in relationship to cos?
Sec(2x) = 1/Cos(2x)
How do you integrate e-2x?
= inegrate (e-2x) / derivate -2x = (e-2x)/-2-> integral esomething = esomething , that's why (e-2x) don't change-> (-2x)' = -2
How do you integrate of e -2x for x0?
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
How you integrate xxxx 1 dx?
.2x^5+x+C
What is the volume of a cube with sides of 2x?
2x x 2x x 2x = 8x3
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
