Ask a New Question

Evaluate the integral:

(cos(2-3x) - tan(5-3x))dx

2 answers

we know that
∫cos(u) du = sin(u)
∫tan(u) du = -log cos(u)

So, what we get here is
(-1/3)sin(2-3x) - (-1/3)(-log cos(5-3x))
= -1/3 (sin(2-3x) + log cos(5-3x)) + C

the correct answer I was given has a negative 1/3 before the log. Also, how did you get the 1/3 value?

Similar Questions

Evaluate the integral: 16csc(x) dx from pi/2 to pi (and determine if it is convergent or divergent).I know how to find the
1. 2 answers
1. Express the given integral as the limit of a Riemann sum but do not evaluate:integral[0 to 3]((x^3 - 6x)dx) 2.Use the
1. 1 answer
Evaluate the integral: 16csc(x) dx from pi/2 to pi (and determine if it is convergent or divergent).I know how to find the
1. 2 answers
a) Let f(z) = z^2 and γ(t) = 1 + it^3, t ∈ [0,1].i) Write out the contour integral ∫γ f(z)dz as an integral with respect
1. 2 answers

more similar questions