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Find the indefinite integral. (Remember to use absolute values where appropriate.)

[tex]\[
\int \csc (7x) \, dx = \square + C
\][/tex]


Sagot :

To find the indefinite integral of [tex]\(\csc(7x)\)[/tex], we need to follow a series of steps. Here's a step-by-step guide:

1. Recognize the Integral Form:
We need to integrate [tex]\(\csc(7x)\)[/tex] with respect to [tex]\(x\)[/tex].

[tex]\[ \int \csc(7x) \, dx \][/tex]

2. Substitute for Simplicity:
Let [tex]\(u = 7x\)[/tex]. Thus, [tex]\(du = 7 \, dx\)[/tex] or [tex]\(\frac{du}{7} = dx\)[/tex].

3. Rewrite the Integral:
Substituting [tex]\(u\)[/tex] and [tex]\(dx\)[/tex] in terms of [tex]\(du\)[/tex]:

[tex]\[ \int \csc(7x) \, dx = \int \csc(u) \cdot \frac{1}{7} \, du \][/tex]

4. Factor out the Constant:
The constant [tex]\(\frac{1}{7}\)[/tex] can be factored out of the integral:

[tex]\[ \frac{1}{7} \int \csc(u) \, du \][/tex]

5. Integrate [tex]\(\csc(u)\)[/tex]:
The integral of [tex]\(\csc(u)\)[/tex] is known:

[tex]\[ \int \csc(u) \, du = \ln \left| \csc(u) - \cot(u) \right| + C \][/tex]

6. Substitute Back [tex]\(u = 7x\)[/tex]:
Replace [tex]\(u\)[/tex] with [tex]\(7x\)[/tex] in the equation:

[tex]\[ \int \csc(7x) \, dx = \frac{1}{7} \ln \left| \csc(7x) - \cot(7x) \right| + C \][/tex]

However, this integral can be simplified differently. Given our earlier completion and the known simplified result, the answer can be presented as:

[tex]\[ \boxed{\int \csc(7x) \, dx = \frac{1}{14} \ln \left| \frac{\cos(7x) - 1}{\cos(7x) + 1} \right| + C} \][/tex]

Thus, the final solution is:

[tex]\[ \boxed{\int \csc(7x) \, dx = \frac{1}{14} \ln \left| \cos(7x) - 1 \right| - \frac{1}{14} \ln \left| \cos(7x) + 1 \right| + C } \][/tex]

Or more compactly:

[tex]\[ \ln \left| \frac{\cos(7x) - 1}{\cos(7x) + 1} \right| / 14 + C \][/tex]

This gives the same result as:

[tex]\[ C + \frac{1}{14} \ln |\cos(7x) - 1| - \frac{1}{14} \ln |\cos(7x) + 1| \][/tex]

Therefore, the integral is:

[tex]\[ \boxed{\log \left| \frac{\cos(7x) - 1}{\cos(7x) + 1} \right| / 14 + C} \][/tex]