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


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:


{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}

Learn more about this topic:

How to Solve Improper Integrals


Chapter 13 / Lesson 13

Improper integrals extend to infinity, typically positively or negatively. Explore the concept of improper integrals and discover the process of solving them using variables to replace infinity and setting a limit.

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