If 2x-1,dx = _e|2x+2x√1+2x|+ x + constant, then - is equal to _______.

Question

If $\displaystyle\int\sqrt{\sec2x-1}\,dx = \alpha\log_e\left|\cos2x+\sqrt{\cos2x}\cdot\sqrt{1+\cos2x}ight|+\beta x + 	ext{constant}$, then $\beta-\alpha$ is equal to \_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

$$\int\sqrt{\sec2x-1}\,dx = \sqrt2\int\frac{\sin x}{\sqrt{2\cos^2x-1}}\,dx$$

Let $\cos x=t$: $-\sqrt2\displaystyle\int\frac{dt}{\sqrt{2t^2-1}} = -\ln\left|\sqrt2\cos x+\sqrt{\cos2x}\right|+c$, which after simplification (writing $2\cos^2x = \cos2x+1$) gives

$$-\frac12\ln\left|\cos2x+\frac12+\sqrt{\cos2x}\sqrt{1+\cos2x}\right|+c$$

matching the given form with $\alpha=-\dfrac12$, $\beta=\dfrac12$, so $\beta-\alpha=1$.

Solutions · 1