Question

evaluate the integral.\
\\(\int \frac{16 \sec^2(4t)}{9 + 4 \tan(4t)} dt\\)\
\\(\int \frac{16 \sec^2(4t)}{9 + 4 \tan(4t)} dt = \square\\)

Explanation:

Step1: Choose Substitution

Let \( u = 9 + 4\tan(4t) \). Then, find \( du \). The derivative of \( \tan(4t) \) with respect to \( t \) is \( 4\sec^2(4t) \) (using the chain rule), so \( du = 4 \cdot 4\sec^2(4t)dt = 16\sec^2(4t)dt \).

Step2: Substitute into Integral

The integral \( \int \frac{16\sec^2(4t)}{9 + 4\tan(4t)}dt \) becomes \( \int \frac{du}{u} \) (since \( 16\sec^2(4t)dt = du \) and \( u = 9 + 4\tan(4t) \)).

Step3: Integrate \( \int \frac{du}{u} \)

The integral of \( \frac{1}{u} \) with respect to \( u \) is \( \ln|u| + C \), where \( C \) is the constant of integration.

Step4: Substitute Back \( u \)

Substitute \( u = 9 + 4\tan(4t) \) back into the result. So we get \( \ln|9 + 4\tan(4t)| + C \).

Answer:

\( \ln|9 + 4\tan(4t)| + C \)