Substituting back, we get tan−1 + C.
- tan−1 (ln x) + C
Recall that = tan−1 u + C. Here we have , and we will need to do
u-substitution. Let u = lnx and du = dx. Substituting into the integrand, we get =
tan−1 u + C. Substituting back, we get tan−1(ln x) + C.
- sin−1 (tan x) + C
Recall that = sin −1 u + C. Here we have , and we will need to
do u-substitution. Let u = tan x and du = sec^2 x dx. Substituting into the integrand, we get.
= sin −1 u + C. Substituting back, we get sin−1 (tan x) + C.
8.
Recall that = sin −1 u + C. Here we have , and we just need to
rearrange the integrand so that it is in the proper form to use the integral formula. If we factor 9
out of the radicand, we get = . Next, we do u-
substitution. Let u = and du = dx. Multiply du by so that du = dx. Substituting into
the integrand, we get = sin−1 u + C. Substituting back, we get sin −1
+ C.
