Q. \[ \int \csc^2(x)\, dx \]

Q: What is \int \csc^2(x)\,dx ?

Use \dfrac{d}{dx}\left(-\cot x\right)=\csc^2 x . So \int \csc^2(x)\,dx=-\cot(x)+C .

Q: How do you derive \dfrac{d}{dx}\left(\cot x\right)=-\csc^2 x ?

Differentiate \cot x=\dfrac{\cos x}{\sin x} using the quotient rule to get -\csc^2 x .

Q: Is there an equivalent form \int \csc^2(x)\,dx using \tan(x/2) ?

Commonly it’s not needed. The direct antiderivative is -\cot x+C . Any substitution should simplify back to -\cot x+C .

Q: What is \int \csc^2(3x)\,dx ?

Let u=3x . Then dx=\dfrac{du}{3} , so \int \csc^2(3x)\,dx=\dfrac{1}{3}\int \csc^2 u\,du=-\dfrac{1}{3}\cot(3x)+C .

Q: What is \int \csc^2(x)\cot(x)\,dx ?

Let u=\cot x , so du=-\csc^2 x\,dx . Then \int \csc^2 x\cot x\,dx=-\int u\,du=-\dfrac{u^2}{2}+C=-\dfrac{1}{2}\cot^2 x+C .

Q: Solve \int \csc^2(x)\,dx definite, e.g. \int_{\pi/4}^{\pi/2}\csc^2 x\,dx ?

Antiderivative: -\cot x . Evaluate: \left[-\cot x\right]_{\pi/4}^{\pi/2} =-\cot(\pi/2)-(-\cot(\pi/4))=0-(-1)=1 .

Answer

We use the identity \( \dfrac{d}{dx}\bigl(\cot(x)\bigr) = -\csc^2(x)\).

So,
\[
\int \csc^2(x)\,dx = -\cot(x) + C.
\]

Detailed Explanation

We want to find an antiderivative of \( \csc^2(x) \). In other words, we want to compute

\[
\int \csc^2(x)\,dx.
\]

Step 1: Recall a key derivative.

A very common identity (and derivative rule) is that

\[
\frac{d}{dx}\bigl(\cot(x)\bigr) = -\csc^2(x).
\]

Step 2: Use that derivative to match the integrand.

Our integrand is \( \csc^2(x) \), but the derivative we know is for \( -\csc^2(x) \). So we rewrite the integrand as the negative of something we can differentiate.

\[
\csc^2(x) = -\bigl(-\csc^2(x)\bigr).
\]

Step 3: Relate the integral to the known derivative.

From

\[
\frac{d}{dx}\bigl(\cot(x)\bigr) = -\csc^2(x),
\]

we can conclude that

\[
\int -\csc^2(x)\,dx = \cot(x) + C.
\]

Step 4: Remove the negative sign correctly.

Because

\[
\int \csc^2(x)\,dx = -\int -\csc^2(x)\,dx,
\]

we get

\[
\int \csc^2(x)\,dx = -\bigl(\cot(x) + C\bigr).
\]

Step 5: Simplify the constant.

A constant with a minus sign is still just a constant, so we can write the final answer as

\[
\int \csc^2(x)\,dx = -\cot(x) + C.
\]

Final Answer:

\[
\boxed{\int \csc^2(x)\,dx = -\cot(x) + C.}
\]

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Calculus FAQ

What is \( \int \csc^2(x)\,dx \) ?

Use \( \dfrac{d}{dx}\left(-\cot x\right)=\csc^2 x \). So \( \int \csc^2(x)\,dx=-\cot(x)+C \).

How do you derive \( \dfrac{d}{dx}\left(\cot x\right)=-\csc^2 x \) ?

Differentiate \( \cot x=\dfrac{\cos x}{\sin x} \) using the quotient rule to get \( -\csc^2 x \).

Is there an equivalent form \( \int \csc^2(x)\,dx \) using \( \tan(x/2) \) ?

Commonly it’s not needed. The direct antiderivative is \( -\cot x+C \). Any substitution should simplify back to \( -\cot x+C \).

What is \( \int \csc^2(3x)\,dx \) ?

Let \( u=3x \). Then \( dx=\dfrac{du}{3} \), so \( \int \csc^2(3x)\,dx=\dfrac{1}{3}\int \csc^2 u\,du=-\dfrac{1}{3}\cot(3x)+C \).

What is \( \int \csc^2(x)\cot(x)\,dx \) ?

Let \( u=\cot x \), so \( du=-\csc^2 x\,dx \). Then \( \int \csc^2 x\cot x\,dx=-\int u\,du=-\dfrac{u^2}{2}+C=-\dfrac{1}{2}\cot^2 x+C \).

Solve \( \int \csc^2(x)\,dx \) definite, e.g. \( \int_{\pi/4}^{\pi/2}\csc^2 x\,dx \) ?

Antiderivative: \( -\cot x \). Evaluate: \( \left[-\cot x\right]_{\pi/4}^{\pi/2} =-\cot(\pi/2)-(-\cot(\pi/4))=0-(-1)=1 \).

Use this to solve ∫csc²(x).
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