# Middle Tennessee State University Integration Question

• January 31, 2022/

### Question Description

Help me study for my Calculus class. I’m stuck and don’t understand.

Two calculus students were asked to evaluate

$\int \mathrm{sin}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}dx.$

Student A arrived at an answer of

$\frac{{\mathrm{sin}}^{2}x}{2}+C$

, while Student B arrived at an answer of

$-\frac{{\mathrm{cos}}^{2}x}{2}+C.\phantom{\rule{0ex}{0ex}}$

Both students received full credit for their responses.

Given that

$\frac{{\mathrm{sin}}^{2}x}{2}\ne -\frac{{\mathrm{cos}}^{2}x}{2}\phantom{\rule{0ex}{0ex}}$

it appears that the students arrived at completely different answers.

Explain why either representation for

$\int \mathrm{sin}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}dx\phantom{\rule{0ex}{0ex}}$

is valid. You may want to consider the graphs of

$y=\frac{{\mathrm{sin}}^{2}x}{2}\phantom{\rule{0ex}{0ex}}$

and

$-\frac{{\mathrm{cos}}^{2}x}{2}\phantom{\rule{0ex}{0ex}}$