Question #40406

This question tests the 'Integration Constant Trap' associated with the Weierstrass substitution $t = \tan(x/2)$.

Because the substitution introduces discontinuities at odd multiples of $\pi$, blindly evaluating the antiderivative $5 \arctan(2 \tan(x/2))$ from $0$ to $\frac{11\pi}{2}$ yields $-5 \arctan(2)$, completely missing the accumulated area.

Furthermore, it traps students who incorrectly assume $\cos x$ has a period of $\pi$ rather than $2\pi$, which would lead them to falsely evaluate the final segment $\int_{5\pi}^{\frac{11\pi}{2}}$ as identical to $\int_0^{\frac{\pi}{2}}$, resulting in the distractor $\frac{25\pi}{2} + 5 \arctan(2)$.