Suppose $X,Y$ are independent Uniform$(0,1)$ random variables. Find the probability $P(Y\geq X\mid Y\geq\dfrac{1}{2})$.
Please note that I know the correct answer and that I have arrived at the correct answer both geometrically and analytically. So that means I solved the problem, which happens to be an easy one. But, I was investigating another method and got a different answer, which, undoubtedly, was wrong. Here is my method:
$$P(Y\geq X\mid Y\geq\frac{1}{2})=\frac{P(Y\geq X,Y\geq\frac {1}{2})}{P(Y\geq\frac{1}{2})}=\dfrac {P(Y\geq \max\{X,\dfrac{1}{2}\})}{\dfrac{1}{2}}.$$ The numerator is to be worked with now. $$P(Y\geq \max\{X,\frac{1}{2}\})=P(Y\geq X\mid X\geq0.5)P(X\geq0.5)+P(Y\geq0.5\mid X\leq0.5)P(X\leq0.5)=P(Y\geq X\mid X\geq0.5)(0.5)+(0.5)(0.5)$$
Now we evaluate $$P(Y\geq X\mid X\geq0.5)=\int_{0.5}^1 \!P(Y\geq k\mid X=k)f_X(k)\,\mathrm{d}k=\int_{0.5}^1\!(1-k)\,\mathrm{d}k=\frac{1}{8}$$
So our final answer turns out to be $$\frac{\dfrac{1}{8}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{2}}{\dfrac{1}{2}}=\dfrac{5}{8}$$
while the correct answer is $\dfrac{3}{4}$.
Your integral calculates $P(Y \geq X \geq 0.5)$, not $P(Y \geq X| X \geq 0.5)$.