Schneider Matthews
04/18/2023 · High School
\( \int \csc \frac { a \phi } { b } \cot \frac { a \phi } { b } d \phi \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the integral
\[
I = \int \csc \left( \frac{a \phi}{b} \right) \cot \left( \frac{a \phi}{b} \right) d\phi,
\]
we can use a substitution to simplify the expression. Let
\[
u = \frac{a \phi}{b}.
\]
Then, the differential \(d\phi\) can be expressed in terms of \(du\):
\[
d\phi = \frac{b}{a} du.
\]
Now, we can rewrite the integral in terms of \(u\):
\[
I = \int \csc(u) \cot(u) \cdot \frac{b}{a} du = \frac{b}{a} \int \csc(u) \cot(u) du.
\]
Next, we need to evaluate the integral
\[
\int \csc(u) \cot(u) du.
\]
The integral of \(\csc(u) \cot(u)\) is a standard integral, and it is known that:
\[
\int \csc(u) \cot(u) du = -\csc(u) + C,
\]
where \(C\) is the constant of integration.
Substituting this back into our expression for \(I\), we have:
\[
I = \frac{b}{a} \left( -\csc(u) + C \right).
\]
Now, substituting back for \(u\):
\[
I = -\frac{b}{a} \csc\left( \frac{a \phi}{b} \right) + C.
\]
Thus, the final result for the integral is:
\[
\int \csc \left( \frac{a \phi}{b} \right) \cot \left( \frac{a \phi}{b} \right) d\phi = -\frac{b}{a} \csc\left( \frac{a \phi}{b} \right) + C.
\]
Quick Answer
The integral \( \int \csc \left( \frac{a \phi}{b} \right) \cot \left( \frac{a \phi}{b} \right) d\phi \) simplifies to \( -\frac{b}{a} \csc\left( \frac{a \phi}{b} \right) + C \).
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