Find the following indefinite integral:

integral tan to the power of 9 x sec squared x d x


a.)
1 over 10 tan to the power of 10 x plus c


b.)
1 over 10 s e c to the power of 10 x plus c


c.)
1 over 9 tan to the power of 10 x plus c


d.)
1 over 9 s e c to the power of 10 x plus c

To solve the integral 

\[ \int \tan^9 x \sec^2 x \, dx, \]

we can use the substitution method. 

Let \( u = \tan x \). Then, we need to find \( du \).

\[ \frac{du}{dx} = \sec^2 x. \]

So,

\[ du = \sec^2 x \, dx. \]

This means \( \sec^2 x \, dx = du \).

Now we can rewrite the integral in terms of \( u \):

\[ \int \tan^9 x \sec^2 x \, dx = \int u^9 \, du. \]

The integral of \( u^9 \) with respect to \( u \) is:

\[ \int u^9 \, du = \frac{u^{10}}{10} + C. \]

Substituting back \( u = \tan x \),

\[ \frac{(\tan x)^{10}}{10} + C. \]

Thus, the answer is

\[ \frac{1}{10} \tan^{10} x + C. \]

This corresponds to option

a.) \( \frac{1}{10} \tan^{10} x + C \).

Would you like further details or have any questions? Here are 5 questions you might want to ask next:

1. How do you integrate functions involving \( \sec x \) and \( \tan x \)?
2. Can you explain the process of integration by parts?
3. What are the common techniques for solving trigonometric integrals?
4. How do you solve integrals involving higher powers of trigonometric functions?
5. Can you provide an example of integrating a product of trigonometric functions?

Tip: Always double-check the substitution step to ensure that all parts of the original integral are correctly transformed.

Learn how to integrate tan^9(x) sec^2(x) step-by-step using the substitution method. This problem covers advanced techniques in integral calculus involving trigonometric functions. Suitable for college-level mathematics students.

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