Question

tan ^ 2 (2x+3)

Answer 1

$\frac{1}{2} \tan(2x+3) - x + C$

Answer 2

To find the integral of $\tan^2(2x+3)$, we use the trigonometric identity $\tan^2\theta = \sec^2\theta - 1$.

Let the given integral be $I$: $I = \int \tan^2(2x+3) dx$

Apply the identity $\tan^2\theta = \sec^2\theta - 1$ with $\theta = 2x+3$: $I = \int (\sec^2(2x+3) - 1) dx$

Now, we can split the integral into two parts: $I = \int \sec^2(2x+3) dx - \int 1 dx$

For the first integral, $\int \sec^2(2x+3) dx$, we use a substitution. Let $u = 2x+3$. Differentiating both sides with respect to $x$: $\frac{du}{dx} = 2$ $du = 2 dx$ $dx = \frac{1}{2} du$

Substitute $u$ and $dx$ into the first integral: $\int \sec^2(u) \left(\frac{1}{2}\right) du = \frac{1}{2} \int \sec^2(u) du$ We know that $\int \sec^2(u) du = \tan(u) + C_1$. So, the first part of the integral is: $\frac{1}{2} \tan(u) + C_1$ Substitute back $u = 2x+3$: $\frac{1}{2} \tan(2x+3) + C_1$

For the second integral, $\int 1 dx$: $\int 1 dx = x + C_2$

Combining both parts: $I = \left(\frac{1}{2} \tan(2x+3) + C_1\right) - (x + C_2)$ $I = \frac{1}{2} \tan(2x+3) - x + (C_1 - C_2)$ Let $C = C_1 - C_2$, which is a new arbitrary constant. $I = \frac{1}{2} \tan(2x+3) - x + C$