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 \(k=2\).

Answer: \(\dfrac34+\mathrm e^4+\dfrac{\mathrm e^8}{4}\).

We start with the main method. Simplify the integrand before integrating, then substitute the upper and lower limits carefully.

First expand the integrand:

\((1+\mathrm e^{2x})^2=1+2\mathrm e^{2x}+\mathrm e^{4x}.\)

So

\(\int_0^2(1+\mathrm e^{2x})^2\,dx=\int_0^2\left(1+2\mathrm e^{2x}+\mathrm e^{4x}\right)\,dx.\)

Integrate term by term:

\(\int\left(1+2\mathrm e^{2x}+\mathrm e^{4x}\right)\,dx =x+\mathrm e^{2x}+\frac{\mathrm e^{4x}}{4}.\)

Now apply the limits:

\(\left[x+\mathrm e^{2x}+\frac{\mathrm e^{4x}}{4}\right]_0^2 =\left(2+\mathrm e^4+\frac{\mathrm e^8}{4}\right)-\left(0+1+\frac14\right).\)

Therefore the exact value is

\(\frac34+\mathrm e^4+\frac{\mathrm e^8}{4}.\)

This completes the solution and gives the required result.

Answer: \(\ln\frac53-\frac{52}{225}\).

We start with the main method. Simplify the integrand before integrating, then substitute the upper and lower limits carefully.

First expand the integrand:

\(\frac{(x-1)^2}{x^3}=\frac{x^2-2x+1}{x^3}=\frac1x-\frac2{x^2}+\frac1{x^3}.\)

So

\(\int_3^5 \frac{(x-1)^2}{x^3}\,\mathrm dx =\int_3^5\left(x^{-1}-2x^{-2}+x^{-3}\right)\mathrm dx.\)

Integrating term by term gives

\(\left[\ln x+\frac2x-\frac1{2x^2}\right]_3^5.\)

At \(x=5\), the value is

\(\ln5+\frac25-\frac1{50}.\)

At \(x=3\), the value is

\(\ln3+\frac23-\frac1{18}.\)

Therefore the exact value is

\(\left(\ln5+\frac25-\frac1{50}\right)-\left(\ln3+\frac23-\frac1{18}\right).\)

The logarithmic part is

\(\ln5-\ln3=\ln\frac53.\)

The numerical part is

\(\frac25-\frac1{50}-\frac23+\frac1{18}=-\frac{52}{225}.\)

Hence

\(\displaystyle \int_3^5 \frac{(x-1)^2}{x^3}\,\mathrm dx=\ln\frac53-\frac{52}{225}.\)

This completes the solution and gives the required result.

Answer: \(\frac{13}{2}+4\ln\frac32\).

We start with the main method. Simplify the integrand before integrating, then substitute the upper and lower limits carefully.

First expand the integrand:

\(\frac{(x+2)^2}{x}=\frac{x^2+4x+4}{x}=x+4+\frac4x.\)

So

\(\int_2^3\frac{(x+2)^2}{x}\,\mathrm dx =\int_2^3\left(x+4+\frac4x\right)\mathrm dx.\)

Integrating term by term gives

\(\left[\frac{x^2}{2}+4x+4\ln x\right]_2^3.\)

At \(x=3\), this is

\(\frac92+12+4\ln3.\)

At \(x=2\), this is

\(2+8+4\ln2.\)

Therefore the exact value is

\(\left(\frac92+12+4\ln3\right)-\left(2+8+4\ln2\right).\)

So

\(\int_2^3\frac{(x+2)^2}{x}\,\mathrm dx =\frac{13}{2}+4\ln\frac32.\)

This completes the solution and gives the required result.

Answer: \(\ln2-\frac14\).

The key point is to integrate each term carefully, then substitute the limits if it is a definite integral.

Integrate term by term:

Therefore

Substitute the limits:

Simplifying gives

\(\ln4-\ln2+\frac14-\frac12.\)

Since \(\ln4-\ln2=\ln2\), the answer is

\(\ln2-\frac14.\)