Find the indefinite integral.

Find the indefinite integral. (Use C for the constant of integration.) csc(x) cot(x) V 16 csc2 X

The Correct Answer and Explanation is:

To solve the indefinite integral of the expression:∫csc⁡(x)cot⁡(x)⋅16csc⁡2(x) dx\int \csc(x) \cot(x) \cdot \sqrt{16 \csc^2(x)} \, dx∫csc(x)cot(x)⋅16csc2(x)​dx

Step 1: Simplify the expression

We start by simplifying the square root term:16csc⁡2(x)=4csc⁡(x)\sqrt{16 \csc^2(x)} = 4 \csc(x)16csc2(x)​=4csc(x)

Thus, the integral becomes:∫csc⁡(x)cot⁡(x)⋅4csc⁡(x) dx\int \csc(x) \cot(x) \cdot 4 \csc(x) \, dx∫csc(x)cot(x)⋅4csc(x)dx

This simplifies further to:∫4csc⁡2(x)cot⁡(x) dx\int 4 \csc^2(x) \cot(x) \, dx∫4csc2(x)cot(x)dx

Step 2: Make a substitution

Notice that the integral contains both csc⁡2(x)\csc^2(x)csc2(x) and cot⁡(x)\cot(x)cot(x), which suggests using substitution. We will use the substitution:u=cot⁡(x)u = \cot(x)u=cot(x)

Thus, the derivative of cot⁡(x)\cot(x)cot(x) with respect to xxx is:dudx=−csc⁡2(x)\frac{du}{dx} = -\csc^2(x)dxdu​=−csc2(x)

or equivalently:du=−csc⁡2(x) dxdu = -\csc^2(x) \, dxdu=−csc2(x)dx

Step 3: Substitute and integrate

Substitute u=cot⁡(x)u = \cot(x)u=cot(x) and du=−csc⁡2(x) dxdu = -\csc^2(x) \, dxdu=−csc2(x)dx into the integral:∫4csc⁡2(x)cot⁡(x) dx=−4∫u du\int 4 \csc^2(x) \cot(x) \, dx = -4 \int u \, du∫4csc2(x)cot(x)dx=−4∫udu

Now, we can integrate with respect to uuu:−4∫u du=−4⋅u22=−2u2-4 \int u \, du = -4 \cdot \frac{u^2}{2} = -2u^2−4∫udu=−4⋅2u2​=−2u2

Step 4: Substitute back

Now, we substitute u=cot⁡(x)u = \cot(x)u=cot(x) back into the expression:−2u2=−2cot⁡2(x)-2u^2 = -2 \cot^2(x)−2u2=−2cot2(x)

Step 5: Add the constant of integration

Finally, since we are evaluating an indefinite integral, we add the constant of integration CCC:∫csc⁡(x)cot⁡(x)⋅16csc⁡2(x) dx=−2cot⁡2(x)+C\int \csc(x) \cot(x) \cdot \sqrt{16 \csc^2(x)} \, dx = -2 \cot^2(x) + C∫csc(x)cot(x)⋅16csc2(x)​dx=−2cot2(x)+C

Final Answer:

−2cot⁡2(x)+C\boxed{-2 \cot^2(x) + C}−2cot2(x)+C​

Explanation:

In this solution, we used substitution to simplify the integral. By recognizing the relationship between the trigonometric functions and using the derivative of cot⁡(x)\cot(x)cot(x), we were able to transform the integral into a more straightforward form, which was then solved using basic integration rules. The constant of integration CCC accounts for the fact that the integral represents a family of functions, all differing by a constant.

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