allery$\int 2x\,\sin(x^2+1)\,dx$ equals
A$\cos(x^2+1)+C$
B$-\dfrac{1}{2}\cos(x^2+1)+C$
C$\sin(x^2+1)+C$
D$-\cos(x^2+1)+C$
Answer & Solution
Correct answer: D. $-\cos(x^2+1)+C$
1. Put $t=x^2+1$, so $dt=2x\,dx$.
2. The integral becomes $\int \sin t\,dt=-\cos t$.
3. Back-substitute: $-\cos(x^2+1)+C$.
4. No extra $\tfrac12$ is needed since the $2x$ is exactly $dt$; option D is the trap.
_Source: syllabus Mathematics Ch 7 "Integrals", p.297_
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