If 2x-1dx= _e| 2x+ + 2x(1+ (1)/( )x)|+ constant, then - is equal to ____________.

Question

If $\int \sqrt{\sec 2x-1}dx=\alpha {\log }_{e}|\cos 2x+\beta +\sqrt{\cos 2x(1+\cos \frac{1}{\beta }x)}|+$ constant, then $\beta -\alpha $ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

$I=\int \sqrt{\sec 2x-1}dx=\int \sqrt{\frac{1-\cos 2x}{\cos 2x}}dx$

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

Let $\sqrt{2\cos }x=t-\sqrt{2}\sin xdx=dt$

$I=\int -\frac{dt}{\sqrt{{t}^{2}-1}}=-\ln |t+\sqrt{{t}^{2}-{a}^{2}}|+c$

$=-\ln |\sqrt{2}\cos x+\sqrt{2{\cos }^{2}}x-1|+c$

$=-\frac{1}{2}\ln |2{\cos }^{2}x+\cos 2x+2\sqrt{2}\sqrt{\cos 2x.{\cos }^{2}x}|+c$

$=-\frac{1}{2}\ln |2\cos 2x+1+2\sqrt{\cos 2x(1+\cos 2x)}|+c$

$=-\frac{1}{2}\ln |\cos 2x+\frac{1}{2}+\sqrt{\cos 2x(1+\cos 2x)}|+c$

$	herefore $ $\alpha =\frac{-1}{2},\beta =\frac{1}{2}$

$	herefore $ $\beta -\alpha =1$

Solutions · 1