Concept:
For integrals of the form:
\[
I=\int_0^a f(x)\,dx
\]
we use the property:
\[
I=\int_0^a f(a-x)\,dx
\]
Step 1: Let:
\[
I=
\int_0^{\frac{\pi}{2}}
\frac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\,dx
\]
Step 2: Use the property:
\[
I=\int_0^{\frac{\pi}{2}} f\left(\frac{\pi}{2}-x\right)\,dx
\]
Since:
\[
\sin\left(\frac{\pi}{2}-x\right)=\cos x
\]
and
\[
\cos\left(\frac{\pi}{2}-x\right)=\sin x
\]
So:
\[
I=
\int_0^{\frac{\pi}{2}}
\frac{\cos^{100}x}{\cos^{100}x+\sin^{100}x}\,dx
\]
Step 3: Add both expressions of \(I\).
\[
2I=
\int_0^{\frac{\pi}{2}}
\left[
\frac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}
+
\frac{\cos^{100}x}{\sin^{100}x+\cos^{100}x}
\right]dx
\]
\[
2I=
\int_0^{\frac{\pi}{2}}1\,dx
\]
\[
2I=\frac{\pi}{2}
\]
Step 4: Divide by \(2\).
\[
I=\frac{\pi}{4}
\]
Therefore,
\[
\boxed{\frac{\pi}{4}}
\]