Question

$\int_{}^{}\frac{\mathbf{\sin}^{\mathbf{3}}\mathbf{2}\mathbf{x}}{\mathbf{\cos}^{\mathbf{5}}\mathbf{2}\mathbf{x}}\mathbf{dx =}$

• A. $\tan^{4}x + c$
• B. $\tan 4x + c$
• C. $\tan^{4}2x + x + c$
• D. $\frac{1}{8}\tan^{4}2x + c$

Answer 1

Answer: (D)

$\frac{1}{8}\tan^{4}2x + c$

Answer 2

Given, $I = \int_{}^{}{\frac{\sin^{3}2x}{\cos^{5}2x}dx}$. The given equation may be written as $\int_{}^{}{\frac{\sin^{3}2x}{\cos^{3}2x}.\frac{1}{\cos^{2}2x}dx} = \int_{}^{}{\tan^{3}2x.\sec^{2}2xdx.}$

Put $\tan 2x = t$ and $2\sec^{2}2xdx = dt$

$I = - \frac{1}{2}\int_{}^{}{t^{3}dt = \frac{t^{4}}{8} + c} = \frac{\tan^{4}2x}{8} + c.$