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Question: Let x<sup>2</sup>¹ np – 1, n Î N then \(\int_{}^{}{x\sqrt{\frac{2\sin(x^{2} + 1)–\sin 2(x^{2} + 1)}{...

Let x2¹ np – 1, n Î N then x2sin(x2+1)sin2(x2+1)2sin(x2+1)+sin2(x2+1)\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 equals

A

ln 12sec(x2+1)\left| \frac{1}{2}\sec(x^{2} + 1) \right| + C

B

ln sec(x2+12)\left| \sec{}\left( \frac{x^{2} + 1}{2} \right) \right| + C

C

12\frac{1}{2}ln | sec(x2 + 1) | + C

D

12\frac{1}{2} ln 2sec(x2+1)\left| \frac{2}{\sec(x^{2} + 1)} \right| + C

Answer

ln sec(x2+12)\left| \sec{}\left( \frac{x^{2} + 1}{2} \right) \right| + C

Explanation

Solution

12\frac{1}{2} dx

Put x2 + 1 = t ̃ 2xdx = dt

= 12\frac{1}{2} 2sintsin2t2sint+sin2t\int_{}^{}\sqrt{\frac{2\sin t–\sin 2t}{2\sin t + \sin 2t}} dt

= 12\frac{1}{2} 22cost2+2cost\int_{}^{}\sqrt{\frac{2–2\cos t}{2 + 2\cos t}} dt

= 12\frac{1}{2} tant2dt\int_{}^{}{\tan\frac{t}{2}dt} = 12lnsect/2+C1/2\frac{\frac{1}{2}\mathcal{l}n|\sec t/2| + C}{1/2}

= ln sec(x2+12)\left| \sec\left( \frac{x^{2} + 1}{2} \right) \right| + C