∫sec 2x tan x dx
sec(2x) or (sec(x))^2 ?
i don't get what you mean?
its sec 2x
i missed something in it sorry. it was ∫sec 2x tan 2x dx
f(x)=sin(2x)/(cos2x)^2
F(x)=sec(2x)/2+C
To solve the integral ∫sec^2(x)tan(x) dx, we can use a method called u-substitution.
Let's make the following substitution:
u = sec(x)
Now, let's find the derivative of u with respect to x:
du/dx = sec(x)tan(x)
To solve for dx, we can rearrange the equation and substitute:
dx = du / (sec(x)tan(x))
Now, let's substitute these values into the integral:
∫sec^2(x)tan(x) dx = ∫u du / (sec(x)tan(x))
Next, we can simplify the integral using the trigonometric identity:
sec(x)tan(x) = sin(x)/cos^2(x)
Substituting this back into our integral, we get:
∫u du / (sin(x)/cos^2(x)) = ∫u du * cos^2(x) / sin(x)
Now, we can cancel out sin(x) and integrate:
∫u du * cos^2(x) = ∫u cos^2(x) du
To integrate ∫u cos^2(x) du, we can use the power rule for integration:
∫u cos^2(x) du = (1/3)u^3 + C
Finally, substituting our original variable back in, we get:
∫sec^2(x)tan(x) dx = (1/3)sec^3(x) + C
Therefore, the integral of sec^2(x)tan(x) dx is (1/3)sec^3(x) + C, where C is the constant of integration.