SOLUTIONS TO PRACTICE PROBLEM SET 23
- ln |tan x| + C
Whenever we have an integral in the form of a quotient, we check to see if the solution is a
logarithm. A clue is whether the numerator is the derivative of the denominator, as it is here.
Let’s use u-substitution. If we let u = tan x, then du = sec^2 x dx. If we substitute into the
integrand, we get dx = . Recall that = |u| + C. Substituting back, we get
dx = ln|tan x| + C.
- −ln |1 − sin x| + C
Whenever we have an integral in the form of a quotient, we check to see if the solution is a
logarithm. A clue is whether the numerator is the derivative of the denominator, as it is here.
Let’s use u-substitution. If we let u = 1 − sin x, then du = − cos x dx. If we substitute into the
integrand, we get dx = − . Recall that = ln|u| + C. Substituting back, we
get dx = − ln|1 − sin x| + C.
- ln |ln x| + C
Whenever we have an integral in the form of a quotient, we check to see if the solution is a
logarithm. A clue is whether the numerator is the derivative of the denominator, as it is here.
Let’s use u-substitution. If we let u = ln x, then du = dx. If we substitute into the integrand,
we get dx = . Recall that = ln| u | + C. Substituting back, we get
dx = ln |ln x | + C.
- −ln |cos x| − x + C
First, let’s rewrite the integrand.
dx = dx = dx = dx − ∫ dx
Whenever we have an integral in the form of a quotient, we check to see if the solution is a
