Understanding the Concept:
Whenever an integral contains a rational expression involving trigonometric functions, it is often helpful to rewrite the numerator in terms of the denominator. This simplifies the expression considerably and makes integration straightforward.
We are given:
\[
I=\int_{0}^{\frac{\pi}{2}} \frac{3\sin x+4\cos x}{\sin x+\cos x}\,dx
\]
Our goal is to simplify the integrand before integrating.
Step 1: Express the numerator suitably.
Observe that:
\[
3\sin x+4\cos x
\]
can be rewritten as:
\[
\frac72(\sin x+\cos x)+\frac12(\cos x-\sin x)
\]
Let us verify:
\[
\frac72\sin x+\frac72\cos x+\frac12\cos x-\frac12\sin x
\]
\[
=\left(\frac72-\frac12\right)\sin x+\left(\frac72+\frac12\right)\cos x
\]
\[
=3\sin x+4\cos x
\]
Hence,
\[
I=\int_{0}^{\frac{\pi}{2}}
\frac{
\frac72(\sin x+\cos x)+\frac12(\cos x-\sin x)
}{
\sin x+\cos x
}\,dx
\]
Splitting the fraction,
\[
I=\int_{0}^{\frac{\pi}{2}}
\left[
\frac72+\frac12\cdot
\frac{\cos x-\sin x}{\sin x+\cos x}
\right]dx
\]
Step 2: Separate the integral.
\[
I=
\frac72\int_{0}^{\frac{\pi}{2}}dx
+
\frac12
\int_{0}^{\frac{\pi}{2}}
\frac{\cos x-\sin x}{\sin x+\cos x}\,dx
\]
Step 3: Evaluate the first integral.
\[
\frac72\int_{0}^{\frac{\pi}{2}}dx
=
\frac72\left[\frac{\pi}{2}\right]
\]
\[
=\frac{7\pi}{4}
\]
Step 4: Evaluate the second integral.
Notice that:
\[
\frac{d}{dx}(\sin x+\cos x)=\cos x-\sin x
\]
Thus,
\[
\int
\frac{\cos x-\sin x}{\sin x+\cos x}\,dx
=
\ln|\sin x+\cos x|
\]
Therefore,
\[
\frac12
\left[
\ln|\sin x+\cos x|
\right]_{0}^{\frac{\pi}{2}}
\]
Substituting limits:
At \(x=\frac{\pi}{2}\),
\[
\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1+0=1
\]
At \(x=0\),
\[
\sin0+\cos0=0+1=1
\]
Hence,
\[
\ln1-\ln1=0
\]
So the second integral contributes:
\[
0
\]
Step 5: Final answer.
Thus,
\[
I=\frac{7\pi}{4}
\]
Hence,
\[
\boxed{\frac{7\pi}{4}}
\]