integral of (1-(sinx)^2))/(cosx)dx
i don't know what to make my "u" for u-substitution
ah
(1-(sinx)^2)) = cos^2 x
Unless I am mistaken or there is a typo all you have is
cos x dx
which is sin x + c
Oh! I didn't realize it was so simple. Thank you!
To determine the appropriate substitution for u, we need to look for a part of the integrand that resembles the derivative of a function that is present in the expression.
In this case, we can observe that the derivative of cos(x) is -sin(x), which is similar to 1 - (sin(x))^2 in the numerator. Therefore, we can choose u = sin(x).
Next, we need to find the derivative of u with respect to x, which will enable us to substitute du into the integral. Taking the derivative of u = sin(x), we get du/dx = cos(x).
Now, we can substitute u = sin(x) and du = cos(x) dx into the integral:
∫(1 - (sin(x))^2)/cos(x) dx
= ∫(1 - u^2)/du [substituting u = sin(x) and du = cos(x) dx]
Now the integral is in terms of u and du. Simplify the integrand:
∫(1 - u^2)/du
= ∫(1/u^2 - 1) du
= ∫(1/u^2) du - ∫(1) du
= -u^(-1) - u + C
Finally, substitute back u = sin(x) to get the final result:
- sin(x)^(-1) - sin(x) + C
Therefore, the integral of (1 - (sin(x))^2)/(cos(x)) dx is -sin(x)^(-1) - sin(x) + C, where C is the constant of integration.