Question:
Evaluate
\[
\int \frac{2\cos x+1}{(2+\cos x)^2}\,dx-\frac{\sin x}{2+\cos x}
\]
Evaluate
\[
\int \frac{2\cos x+1}{(2+\cos x)^2}\,dx-\frac{\sin x}{2+\cos x}
\]
Show Hint
Before integrating, check whether the integrand is directly the derivative of a given expression. This can simplify the problem immediately.
Updated On: Jun 26, 2026
- \(\frac{1}{2+\cos x}+C\)
- \(\sin x+C\)
- \(\frac{2}{2+\cos x}+C\)
- \(C\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Observe the derivative of the second term.
Consider \[ \frac{\sin x}{2+\cos x} \] Differentiate it: \[ \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{(2+\cos x)\cos x-\sin x(-\sin x)}{(2+\cos x)^2} \] \[ = \frac{2\cos x+\cos^2x+\sin^2x}{(2+\cos x)^2} \] Using \[ \sin^2x+\cos^2x=1, \] we get \[ \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{2\cos x+1}{(2+\cos x)^2} \]
Step 2: Integrate the given integrand.
Thus, \[ \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx = \frac{\sin x}{2+\cos x}+C \]
Step 3: Substitute in the given expression.
Now, \[ \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx-\frac{\sin x}{2+\cos x} \] \[ = \frac{\sin x}{2+\cos x}+C-\frac{\sin x}{2+\cos x} \] \[ =C \]
Step 4: Final conclusion.
Hence, \[ \boxed{C} \]
Consider \[ \frac{\sin x}{2+\cos x} \] Differentiate it: \[ \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{(2+\cos x)\cos x-\sin x(-\sin x)}{(2+\cos x)^2} \] \[ = \frac{2\cos x+\cos^2x+\sin^2x}{(2+\cos x)^2} \] Using \[ \sin^2x+\cos^2x=1, \] we get \[ \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{2\cos x+1}{(2+\cos x)^2} \]
Step 2: Integrate the given integrand.
Thus, \[ \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx = \frac{\sin x}{2+\cos x}+C \]
Step 3: Substitute in the given expression.
Now, \[ \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx-\frac{\sin x}{2+\cos x} \] \[ = \frac{\sin x}{2+\cos x}+C-\frac{\sin x}{2+\cos x} \] \[ =C \]
Step 4: Final conclusion.
Hence, \[ \boxed{C} \]
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