Question

In Exercises \(25-46,\) use substitution to evaluate the integral. $$\int \frac{\sin (2 t+1)}{\cos ^{2}(2 t+1)} d t$$

Short answer

\(\frac{1}{2} \tan(2t + 1) - \frac{1}{2} (2t+1) + C \)

Step-by-step solution

Set the substitution

Let \( u = 2t + 1 \). Then, \( du = 2dt \), so \( dt = \frac{1}{2} du \).

Substitute back into the integral

Using the substitution, the original integral becomes \( \int \frac{\sin (u)}{\cos ^{2}(u)} du \). This can be simplified using trigonometric identity: \( \sin(u) = \frac{1-\cos^2(u)}{cos(u)} \) yielding \( \int \frac{1-\cos^2(u)}{\cos(u)} \times \frac{1}{2} du = \frac{1}{2} \int \frac{1}{\cos(u)} du - \frac{1}{2} \int du \).

Solve the Integral

The integral can be solved as \( \frac{1}{2} \int\frac{1}{\cos(u)} du - \frac{1}{2} \int du = \frac{1}{2} \tan(u) - \frac{1}{2} u + C \), where C is the constant of integration.

Replace the original variable

Now replace \(u\) with original variable \( t \). The final answer will be \( \frac{1}{2} \tan(2t + 1) - \frac{1}{2} (2t+1) + C \).