Question

Evaluate: \[ \int_{\pi/2}^{4051\pi/2}\frac{\cos^22x}{1+\sin2x}\,dx \]

• A. \(2026\pi\)
• B. \(2047\pi\)
• C. \(2027\pi\)
• D. \(2025\pi\)

Hint:

Whenever expressions like \[ \frac{\cos^2\theta}{1+\sin\theta} \] appear, multiply identities cleverly to simplify before integrating.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand.
Since $\cos^2 2x=(1-\sin2x)(1+\sin2x)$, dividing by $1+\sin2x$ leaves \[ \frac{\cos^2 2x}{1+\sin2x}=1-\sin2x. \]
Step 2: Rewrite the integral.
\[ I=\int_{\pi/2}^{4051\pi/2}(1-\sin2x)\,dx. \]
Step 3: Integrate.
\[ I=\left[x+\frac{\cos2x}{2}\right]_{\pi/2}^{4051\pi/2}. \]
Step 4: Work out the $x$ part.
\[ \frac{4051\pi}{2}-\frac{\pi}{2}=\frac{4050\pi}{2}=2025\pi. \]
Step 5: Work out the cosine part.
$\cos(4051\pi)=(-1)^{4051}=-1$ and $\cos\pi=-1$, so $\dfrac{-1}{2}-\dfrac{-1}{2}=0$. The cosine terms cancel.
Step 6: Combine.
\[ I=2025\pi+0=2025\pi. \] \[ \boxed{2025\pi} \]