Question

$\int_{}^{}{\sqrt{\mathbf{2 +}\mathbf{\sin}\mathbf{3}\mathbf{x}}\mathbf{.}\mathbf{\cos}\mathbf{3}\mathbf{xdx}}\mathbf{=}$

• A. $\frac{2}{9}\sqrt{2 + \sin 3x} + c$
• B. $\frac{2}{3}(2 + \sin 3x)^{2/3} + c$
• C. $\frac{2}{3}(2 + \sin 3x)^{3/2} + c$
• D. $\frac{2}{9}(2 + \sin 3x)^{3/2} + C$

Answer 1

Answer: (D)

$\frac{2}{9}(2 + \sin 3x)^{3/2} + C$

Answer 2

Put $(2 + \sin 3x) = t$ $\Rightarrow 3\cos 3xdx = dt$

$\int_{}^{}{\sqrt{2 + \sin 3x}.\cos 3xdx = \frac{1}{3}\int_{}^{}{\sqrt{t}dt}} = \frac{1}{3}.\frac{t^{3/2}}{3/2} + c$ $= \frac{2}{9}.(2 + \sin 3x)^{3/2} + C$.