Evaulate:
integral 3x (sinx/cos^4x) dx
I think it's sec3 x , but that from using a piece of software, so you'll have to verify that.
Using uppercase 's' for the integral sign we have
S 3sin(x)/cos4dx or
S cos-4(x)*3sin(x)dx
If you let u = cos(x) then du = -sin(x)dx
So the integral becomes
-3 S u-4du = -3 * (1/-3)u-3 = 1/cos3 x
sec3 x
To evaluate the integral ∫ 3x(sin(x)/cos^4(x)) dx, you correctly identified that it simplifies to ∫ 3sin(x)/cos^4(x) dx. Let's go through the steps to evaluate this integral.
1. Rewrite the integral in a convenient form: ∫ cos^(-4)(x) * 3sin(x) dx.
2. Use a substitution to simplify the integral. Let u = cos(x), then du = -sin(x) dx.
3. Substitute u and du into the integral: ∫ -3u^(-4) du.
4. Integrate the simplified expression: ∫ -3u^(-4) du = -3 * (u^(-3) / -3) + C, where C is the constant of integration.
5. Simplify the expression: -3 * (u^(-3) / -3) + C = 1/u^3 + C.
6. Substitute back u = cos(x): 1/cos^3(x) + C.
7. Simplify further using the identity sec^3(x) = 1/cos^3(x): sec^3(x) + C.
Therefore, the evaluated integral is sec^3(x) + C.