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

Concept:
• Use substitution: $u = \sin t - \cos t$

Step 1: Let
\[ u = \sin t - \cos t \Rightarrow \frac{du}{dt} = \cos t + \sin t \] \[ du = (\sin t + \cos t)dt \]

Step 2: Rewrite denominator
\[ 13 + 36\sin^2 t = 13 + 18(1 - \cos2t) \] Using identity simplification gives: \[ = 49 - 36(\sin t - \cos t)^2 = 49 - 36u^2 \]

Step 3: Substitute
\[ \int \frac{du}{49 - 36u^2} \]

Step 4: Use standard form
\[ \int \frac{du}{a^2 - b^2u^2} = \frac{1}{2ab}\log\left|\frac{a+bu}{a-bu}\right| \] Here $a=7,\; b=6$ \[ = \frac{1}{84}\log\left|\frac{7+6u}{7-6u}\right| + C \]

Step 5: Back substitute
\[ = \frac{1}{84}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C \] Final Conclusion:
Option (A)