Question

$\displaystyle \int \frac{\sin t + \cos t}{13 + 36\sin^2 t} \, dt$ is equal to:

• A. $\frac{1}{84}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$
• B. $\frac{1}{81}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$
• C. $\frac{1}{84}\log\left|\frac{7 - 6(\sin t - \cos t)}{7 + 6(\sin t - \cos t)}\right| + C$
• D. $\frac{1}{48}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$
• E. $\frac{1}{64}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$

Hint:

Try substitution combining $\sin t$ and $\cos t$ when they appear together.

Answer 1

The Correct Option is A