Question:
Evaluate:
\[
\int_{\pi/2}^{4051\pi/2}\frac{\cos^22x}{1+\sin2x}\,dx
\]
Evaluate:
\[
\int_{\pi/2}^{4051\pi/2}\frac{\cos^22x}{1+\sin2x}\,dx
\]
Show Hint
Whenever expressions like
\[
\frac{\cos^2\theta}{1+\sin\theta}
\]
appear, multiply identities cleverly to simplify before integrating.
Updated On: Jun 17, 2026
- \(2026\pi\)
- \(2047\pi\)
- \(2027\pi\)
- \(2025\pi\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Use trigonometric identities to simplify the integrand.
Step 1: Simplify the fraction. Using \[ \cos^2\theta=(1-\sin\theta)(1+\sin\theta) \] with \(\theta=2x\), \[ \cos^22x=(1-\sin2x)(1+\sin2x) \] Hence, \[ \frac{\cos^22x}{1+\sin2x} = 1-\sin2x \] Therefore, \[ I=\int_{\pi/2}^{4051\pi/2}(1-\sin2x)\,dx \]
Step 2: Integrate. \[ I= \left[x+\frac{\cos2x}{2}\right]_{\pi/2}^{4051\pi/2} \] Now, \[ x\text{-part}= \frac{4051\pi}{2}-\frac{\pi}{2} = 2025\pi \] Next, \[ \cos(4051\pi)=(-1)^{4051}=-1 \] and \[ \cos\pi=-1 \] Hence cosine terms cancel: \[ \frac{-1}{2}-\frac{-1}{2}=0 \] Therefore, \[ I=2025\pi \] Hence, \[ \boxed{2025\pi} \]
Step 1: Simplify the fraction. Using \[ \cos^2\theta=(1-\sin\theta)(1+\sin\theta) \] with \(\theta=2x\), \[ \cos^22x=(1-\sin2x)(1+\sin2x) \] Hence, \[ \frac{\cos^22x}{1+\sin2x} = 1-\sin2x \] Therefore, \[ I=\int_{\pi/2}^{4051\pi/2}(1-\sin2x)\,dx \]
Step 2: Integrate. \[ I= \left[x+\frac{\cos2x}{2}\right]_{\pi/2}^{4051\pi/2} \] Now, \[ x\text{-part}= \frac{4051\pi}{2}-\frac{\pi}{2} = 2025\pi \] Next, \[ \cos(4051\pi)=(-1)^{4051}=-1 \] and \[ \cos\pi=-1 \] Hence cosine terms cancel: \[ \frac{-1}{2}-\frac{-1}{2}=0 \] Therefore, \[ I=2025\pi \] Hence, \[ \boxed{2025\pi} \]
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