Integrate $\int \sin (3 x) e^{\cos (3 x)} d x$ .
A) $-e^{\cos (3 x)}+C$
B) $e^{\cos (3 x)}+C$
C) $-\frac{1}{3} e^{\cos (3 x)}+C$
D) $\frac{1}{3} e^{\cos (3 x)}+C$

Question

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = \cos(3x) $, then $ du = -3\sin(3x)dx $, or $ \sin(3x)dx = -\frac{1}{3}du $. The integral becomes $-\frac{1}{3}\int e^u du = -\frac{1}{3} e^u + C = -\frac{1}{3} e^{\cos(3x)} + C$.

Integrate ∫sin⁡(3x)ecos⁡(3x)dx\int \sin (3 x) e^{\cos (3 x)} d x∫sin(3x)ecos(3x)dx A)

Integrate $\int \sin (3 x) e^{\cos (3 x)} d x$
A)