Find the intergral of 1/[cos(^2) x
What is the derivative of tan x ? (do it as a u/v,or sinx/cosx)
remember that cos^2+sin^2x is 1.
Can you please explain how to work it because I still cannot understand.
Since the derivative of tan x is sec^2 x (which is 1/cos^2 x) , the integral of 1/cos^2 x must be tan x. That is what BobPursley was saying.
To find the integral of 1/[cos(^2) x], you can use a trigonometric identity and then apply a substitution.
We know that the identity 1 + tan(^2) x = sec(^2) x holds true for all values of x. Rearranging this identity gives us tan(^2) x = sec(^2) x - 1.
Now let's substitute tan(^2) x = sec(^2) x - 1 into the integral:
∫ 1/[cos(^2) x] dx = ∫ 1/(sec(^2) x - 1) dx
Next, we can apply a substitution. Let u = tan(x). Then du = sec(^2) x dx.
Now let's rewrite the integral in terms of u:
∫ 1/(sec(^2) x - 1) dx = ∫ 1/(u^2 + 1) du
The integral on the right side is a well-known integral that can be solved by using the inverse tangent function:
∫ 1/(u^2 + 1) du = arctan(u) + C
Finally, substitute back u = tan(x) to get the original variable:
= arctan(tan(x)) + C
Therefore, the integral of 1/[cos(^2) x] is arctan(tan(x)) + C, where C is the constant of integration.