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What's the integral of csc(x)cot(x)?

Let’s enhance the clarity and readability of the content while maintaining the mathematical expressions.

The integral of csc(x)cot(x)\csc(x) \cot(x) can be expressed as:

csc(x)cot(x)dx=csc(x)+C\int \csc(x) \cot(x) \, dx = -\csc(x) + C

To compute this integral, we can use the substitution u=csc(x)u = \csc(x). Consequently, the derivative dudx=csc(x)cot(x)\frac{du}{dx} = -\csc(x) \cot(x). This allows us to rewrite the integral as follows:

csc(x)cot(x)dx=du\int \csc(x) \cot(x) \, dx = -\int du

Next, we can utilize the trigonometric identity csc(x)=1sin(x)\csc(x) = \frac{1}{\sin(x)} to express the integral in another form:

csc(x)cot(x)dx=(cos(x)sin2(x))dx\int \csc(x) \cot(x) \, dx = \int \left( \frac{\cos(x)}{\sin^2(x)} \right) \, dx

Now, we apply another substitution, letting u=sin(x)u = \sin(x). Thus, we have dudx=cos(x)\frac{du}{dx} = \cos(x), which transforms our integral into:

(cos(x)sin2(x))dx=duu2\int \left( \frac{\cos(x)}{\sin^2(x)} \right) \, dx = -\int \frac{du}{u^2}

Integrating the expression duu2-\int \frac{du}{u^2} yields:

duu2=(u1)+C=1u+C=csc(x)+C-\int \frac{du}{u^2} = -(-u^{-1}) + C = \frac{1}{u} + C = -\csc(x) + C

Thus, we arrive at the final result for the integral:

csc(x)cot(x)dx=csc(x)+C\int \csc(x) \cot(x) \, dx = -\csc(x) + C
Answered by: Prof. Isabella Taylor
A-Level Maths Tutor
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