Answer: \(k=2\).

Expand both squared expressions:

\((\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta,\)

and

\((\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta.\)

Adding these, the middle terms cancel:

\((\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\sin^2\theta+2\cos^2\theta.\)

Using \(\sin^2\theta+\cos^2\theta=1\), this becomes

\(2.\)

Therefore

\(\displaystyle \int_{\frac\pi2}^{\frac{3\pi}{2}}\left((\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2\right)\,\mathrm d\theta =\int_{\frac\pi2}^{\frac{3\pi}{2}}2\,\mathrm d\theta.\)

So

\(\displaystyle \int_{\frac\pi2}^{\frac{3\pi}{2}}2\,\mathrm d\theta =2\left[\theta\right]_{\frac\pi2}^{\frac{3\pi}{2}} =2\left(\frac{3\pi}{2}-\frac\pi2\right)=2\pi.\)

Hence \(k=2\).