Question

The value of $\int \frac{\cos x}{\sin^2 x} \, dx$ is given by}

• A. $-\csc x + C$
• B. $-\csc x \cot x + C$
• C. $\csc x + C$
• D. $\csc x \cot x + C$

Hint:

Recognizing standard derivative forms like $\frac{d}{dx}(\csc x) = -\csc x \cot x$ helps solve such trigonometric integrals instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Trigonometric integrals can be simplified by breaking them down into products of basic reciprocal trigonometric functions.

Step 2: Detailed Explanation:
Let us rewrite the given integral:
\[ I = \int \frac{\cos x}{\sin^2 x} \, dx \]
Separate the fraction into two trigonometric factors:
\[ I = \int \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} \, dx \]
Using standard trigonometric identities ($\frac{1}{\sin x} = \csc x$ and $\frac{\cos x}{\sin x} = \cot x$):
\[ I = \int \csc x \cdot \cot x \, dx \]
We know from standard differentiation rules that:
\[ \frac{d}{dx}(\csc x) = -\csc x \cot x \implies \int \csc x \cot x \, dx = -\csc x + C \]
Therefore, the value of the integral is $-\csc x + C$.

Step 3: Final Answer
The correct option is (A).