Question

Evaluate
\[ \int \frac{e^{\cot x}}{\sin^2 x}\left(2\log\cosec x+\sin2x\right)\,dx \]

• A. \(-2e^{\cot x}\log(\cosec^2x)+C\)
• B. \(-2e^{\cot x}\log(\cosec x)+C\)
• C. \(-2e^{\cot x}\log(\cosec x+\sin x)+C\)
• D. \(-2e^{\cot x}\log(\cosec x-\cot x)+C\)

Hint:

When the integrand contains \(e^{\cot x}\), remember that \(\frac{d}{dx}(e^{\cot x})=-e^{\cot x}\cosec^2x\).

Answer 1

The Correct Option is B

Step 1: Consider the option form.
Let
\[ F(x)=-2e^{\cot x}\log(\cosec x) \]

Step 2: Differentiate \(F(x)\).
Using product rule,
\[ F'(x)=-2\left[e^{\cot x}(-\cosec^2x)\log(\cosec x)+e^{\cot x}(-\cot x)\right] \]
\[ F'(x)=2e^{\cot x}\cosec^2x\log(\cosec x)+2e^{\cot x}\cot x \]

Step 3: Rewrite in the required form.
Since
\[ \cosec^2x=\frac1{\sin^2x} \] and
\[ 2\cot x=\frac{2\cos x}{\sin x} =\frac{\sin2x}{\sin^2x}, \]
we get
\[ F'(x)=\frac{e^{\cot x}}{\sin^2x}\left(2\log\cosec x+\sin2x\right) \]

Step 4: Final conclusion.
Therefore,
\[ \int \frac{e^{\cot x}}{\sin^2 x}\left(2\log\cosec x+\sin2x\right)\,dx = \boxed{-2e^{\cot x}\log(\cosec x)+C} \]