Question

Let x² ¹ np –1, n Î N, then $\int_{}^{}{x\sqrt{\frac{2\sin(x^{2} + 1)–\sin 2(x^{2} + 1)}{2\sin(x^{2} + 1) + \sin 2(x^{2} + 1)}}}$dx

is equal to:

• A. ln$\left| \frac{1}{2}\sec(x^{2} + 1) \right|$ + C
• B. ln$\left| \sec\left( \frac{x^{2} + 1}{2} \right) \right|$ + C
• C. $\frac{1}{2}$ln |sec(x² + 1)| + C
• D. $\frac{1}{2}$ln $\left| \frac{2}{\sec(x^{2} + 1)} \right|$ + C

Answer 1

Answer: (B)

ln$\left| \sec\left( \frac{x^{2} + 1}{2} \right) \right|$ + C

Answer 2

= $\frac { 1 } { 2 }$ $\int_{}^{}{2x\sqrt{\frac{2\sin(x^{2} + 1)–\sin 2(x^{2} + 1)}{2\sin(x^{2} + 1) + \sin 2(x^{2} + 1)}}}$dx

x² + 1 = t ̃ 2x dx = dt

I = $\frac { 1 } { 2 }$ $\int_{}^{}\sqrt{\frac{2\sin t–\sin 2t}{2\sin t + \sin 2t}}$dt

= $\frac { 1 } { 2 }$ $\int_{}^{}\sqrt{\frac{2–2\cos t}{2 + 2\cos t}}$dt = $\frac{1}{2}$ $\int_{}^{}{\tan\frac{t}{2}}$dt

= $\frac { 1 } { 2 }$ $\frac{\mathcal{l}n\left| \sec\frac{t}{2} \right|}{\frac{1}{2}}$ + c = ln $\left| \sec\left( \frac{x^{2} + 1}{2} \right) \right|$ + c