Evaluate the indefinite integral by using the given substitution to reduce the integral to standard form.
\[
\int 8 \sec ^{2}(4 x) \tan (4 x) d x
\]
a. $u=\tan (4 x)$
b. $u=\sec (4 x)$
a. Using $u=\tan (4 x), \int 8 \sec ^{2}(4 x) \tan (4 x) d x=$
Answer
So, the solution to the integral $\int 8 \sec^2(4x) \tan(4x) dx$ is $\boxed{\tan^2(4x) + C}$.
Step 1 ：First, we use the given substitution $u = \tan(4x)$, then the derivative of $u$ with respect to $x$ is $du/dx = 4\sec^2(4x)$, or $dx = du / (4\sec^2(4x))$.
Step 2 ：Substitute $u$ and $dx$ into the integral, we get $\int 8 \sec^2(4x) u du / (4\sec^2(4x))$.
Step 3 ：Simplify the integral, we get $\int 2u du$.
Step 4 ：Integrate $2u$ with respect to $u$, we get $u^2 + C$, where $C$ is the constant of integration.
Step 5 ：Substitute $u = \tan(4x)$ back into the integral, we get $\tan^2(4x) + C$.
Step 6 ：So, the solution to the integral $\int 8 \sec^2(4x) \tan(4x) dx$ is $\boxed{\tan^2(4x) + C}$.
