# Find the indefinite integral. integral: square root tan {eq}(7x) * (sec(7x))^2 dx {/eq}

## Question:

Find the indefinite integral. integral: square root tan {eq}(7x) * (sec(7x))^2 dx {/eq}

## Integration Using Substitution:

For evaluating an integration where trigonometric function is there, we will substitute the trigonometric function with a variable. We will evaluate the integration with this variable.

We will use the following integration rule to solve this problem:

1.)

{eq}\int x^n \, dx = \frac{x^{(n+1)}}{n+1} + C {/eq}

Where {eq}C {/eq} is an integral constant.

## Answer and Explanation: 1

The integral is given by:

{eq}\int \sqrt{ \tan (7x)} \sec^2 (7x) \, dx {/eq}

Let us assume that:

{eq}v = \tan (7x) \\ \Rightarrow dv = 7 \sec^2 (7x) \, dx \\ \Rightarrow \sec^2 (7x) \, dx = \frac{dv}{7} {/eq}

Substitute {eq}v {/eq} into integration and we have:

{eq}\begin{align*} \int \sqrt{ \tan (7x)} \sec^2 (7x) \, dx &= \frac{1}{7} \int \sqrt{v} \, dv \\ &= \frac{1}{7} \int v^{\frac{1}{2}} \, dv \\ &= \frac{1}{7} \left[ \frac{v^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} \right] + C \hspace{1 cm} \left[ \text{ By using [1] } \right] \\ &= \frac{2}{21} v^{\frac{3}{2}} + C \end{align*} {/eq}

Undo substitution and we have:

{eq}\begin{align*} \int \sqrt{\tan (7x)} \sec^2 (7x) \, dx &= \frac{2}{21} (\tan (7x))^{\frac{3}{2}} + C \\ &= \frac{2}{21} \tan (7x) \sqrt{\tan (7x)} + C \end{align*} {/eq}