Evaluate the integral by making a substitution and then using a table of integrals.
-$$\int \frac { \cos \theta } { \sin \theta \sqrt { 49 + \sin ^ { 2 } \theta } } d \theta$$
A) $- \frac { 1 } { 7 } \ln \left| \frac { 7 + \sqrt { 49 + \sin ^ { 2 } \theta } } { \sin \theta } \right| + C$
B) $- \frac { 1 } { 7 } \ln \left| \frac { 7 \cos \theta + \sqrt { 49 + \sin ^ { 2 } \theta } } { \sin \theta } \right| + C$
C) $- \frac { 1 } { 7 } \ln \left| \frac { 7 - \sqrt { 49 + \sin ^ { 2 } \theta } } { \sin \theta } \right| + C$
D) $- \frac { 1 } { 7 } \ln \left| \frac { 7 + \sqrt { 49 + \sin ^ { 2 } \theta } } { \sin \theta \cos \theta } \right| + C$

Question

Quizplus · Accepted Answer

To solve the integral, let $u = \sqrt{49 + \sin^2\theta}$. Differentiating both sides with respect to $\theta$ gives $du = \frac{\sin\theta\cos\theta}{\sqrt{49 + \sin^2\theta}} d\theta$, which simplifies the integral into a form that can be directly integrated using a table of integrals, leading to the solution provided in option A.

Evaluate the Integral by Making a Substitution and Then Using ∫cos⁡θsin⁡θ49+sin⁡2θdθ\int \frac { \cos \theta } { \sin \theta \sqrt { 49 + \sin ^ { 2 } \theta } } d \theta∫sinθ49+sin2θ​cosθ​dθ

Evaluate the Integral by Making a Substitution and Then Using $\int \frac { \cos \theta } { \sin \theta \sqrt { 49 + \sin ^ { 2 } \theta } } d \theta$