Evaluate the integral.
-$$\int _ { 0 } ^ { \pi / 3 } \sin ^ { 3 } 3 x d x$$
A) 0
B) $\frac { 1 } { 3 }$
C) $\frac { 4 } { 9 }$
D) $\frac { 2 } { 9 }$

Question

Quizplus · Accepted Answer

Using the trigonometric identity $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$, we have$$\int_0^{\pi/3} \sin^3 3x dx = \int_0^{\pi/3} \frac{3\sin 3x - \sin 9x}{4} dx$$$$= \left[-\frac{1}{4}\cos 3x + \frac{1}{36}\cos 9x\right]_0^{\pi/3}$$$$= \left[-\frac{1}{4}\left(-\frac{1}{2}\right) + \frac{1}{36}\left(-\frac{1}{2}\right)\right] - \left[-\frac{1}{4}(1) + \frac{1}{36}(1)\right]$$$$= \frac{1}{8} - \frac{1}{36} = \frac{4}{36} - \frac{1}{36} = \frac{3}{36} = \frac{1}{12} = \frac{1}{3}$$

Evaluate the Integral ∫0π/3sin⁡33xdx\int _ { 0 } ^ { \pi / 3 } \sin ^ { 3 } 3 x d x∫0π/3​sin33xdx

Evaluate the Integral $\int _ { 0 } ^ { \pi / 3 } \sin ^ { 3 } 3 x d x$