Question

I = $\int_{}^{}\left( \frac{1 + \cos 2x}{2} \right)^{2}$dx

= $\frac { 1 } { 4 }$ $\int_{}^{}{(1 + \cos^{2}2x + 2\cos 2x)}$dx

= $\frac { 1 } { 4 }$ $\int_{}^{}\left( 1 + \frac{1 + \cos 4x}{2} + 2\cos 2x \right)$dx

= $\frac { 1 } { 8 }$ $\int_{}^{}{(3 + \cos 4x + 4\cos 2x)dx}$

= $\frac{3x}{8}$+ $\frac{\sin 4x}{32}$+ $\frac{4\sin 2x}{8 \times 2}$

• A. $\frac{3}{n}$ln$\left( \frac{x^{n}}{x^{n} + 1} \right)$
• B. $\frac{1}{n}$ln $\left( \frac{x^{n}}{x^{n} + 1} \right)$
• C. $\frac{3}{n}$ln$\left( \frac{x^{n} + 1}{x^{n}} \right)$
• D. $\frac{\mathbf{3}}{\mathbf{n}}$ ln $\left( \frac{\mathbf{x}^{\mathbf{n}}\mathbf{+ 1}}{\mathbf{x}^{\mathbf{n}}} \right)$

Answer 1

Answer: (A)

$\frac{3}{n}$ln$\left( \frac{x^{n}}{x^{n} + 1} \right)$

Answer 2

$\int \frac { d x ^ { 3 } } { x ^ { 3 } \left( x ^ { n } + 1 \right) }$ = 3 $\int_{}^{}\frac{x^{2}dx}{x^{3}(x^{n} + 1)}$= 3 $\int_{}^{}\frac{dx}{x(x^{n} + 1)}$

= 3dx = $\int_{}^{}\frac{dt}{t(t + 1)}$

= $\int_{}^{}\frac{(t + 1) - t}{t(t + 1)}$dt =$\int_{}^{}\left( \frac{1}{t} - \frac{1}{t + 1} \right)$dt

=$\frac{3}{n}$ (lnt – ln (t + 1)) + c

=$\frac{3}{n}$ln $\left( \frac{t}{t + 1} \right)$+ C