Question

If $\int_{}^{}{\sin^{3}x\mspace{6mu}.\mspace{6mu}\cos x\mspace{6mu} dx =}$, then $\frac{\sin^{4}x\cos^{2}x}{8} + c$

• A. $4\sin^{4}x + c$
• B. $\int_{}^{}{a^{3x + 3}dx} =$
• C. $\frac{a^{3x + 3}}{\log a} + c$
• D. $\frac{a^{3x + 3}}{3\log a} + c$

Answer 1

Answer: (C)

$\frac{a^{3x + 3}}{\log a} + c$

Answer 2

Given that

$\frac{1}{2}\log\left( {logtan}\frac{x}{2} \right) + c$

$\int_{}^{}{\frac{1}{\cos^{2}x(1 - \tan x)^{2}}dx =}$

$\frac{1}{\tan x - 1} + c$

$\frac{1}{1 - \tan x} + c$

$- \frac{1}{3}\frac{1}{(1 - \tan x)^{3}} + c$