Question

If $\frac{5\pi}{4}<x<\frac{7\pi}{4}$, then $\int \sqrt{\frac{1-\sin 2x}{1+\sin 2x}} dx =$

• A. $-\sec^2(\frac{\pi}{4}-x)+c$
• B. $-\log\sec(\frac{\pi}{4}-x)+c$
• C. $\sec^2(\frac{\pi}{4}-x)+c$
• D. $\log\sec(\frac{\pi}{4}-x)+c$

Hint:

When simplifying expressions like $\sqrt{1 \pm \sin 2x}$, use the perfect square identities $(\cos x \pm \sin x)^2$. Always be careful with the absolute value that results from the square root and determine the correct sign based on the given interval.

Answer 1

The Correct Option is D

First, we simplify the expression inside the square root.
Use the identities $1 = \cos^2 x + \sin^2 x$ and $\sin 2x = 2\sin x \cos x$.
$1-\sin 2x = \cos^2 x + \sin^2 x - 2\sin x \cos x = (\cos x - \sin x)^2$.
$1+\sin 2x = \cos^2 x + \sin^2 x + 2\sin x \cos x = (\cos x + \sin x)^2$.
The integrand becomes $\sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} = \left|\frac{\cos x - \sin x}{\cos x + \sin x}\right|$.
Divide the numerator and denominator inside the absolute value by $\cos x$:
$\left|\frac{1 - \tan x}{1 + \tan x}\right| = |\tan(\frac{\pi}{4}-x)|$.
Now we determine the sign of $\tan(\frac{\pi}{4}-x)$ in the given interval $\frac{5\pi}{4}<x<\frac{7\pi}{4}$.
$-\frac{7\pi}{4}<-x<-\frac{5\pi}{4}$.
$\frac{\pi}{4} - \frac{7\pi}{4}<\frac{\pi}{4} - x<\frac{\pi}{4} - \frac{5\pi}{4}$.
$-\frac{6\pi}{4}<\frac{\pi}{4} - x<-\frac{4\pi}{4} \implies -\frac{3\pi}{2}<\frac{\pi}{4} - x<-\pi$.
This interval for $(\frac{\pi}{4}-x)$ is in the second quadrant, where the tangent function is negative.
So, $|\tan(\frac{\pi}{4}-x)| = -\tan(\frac{\pi}{4}-x)$.
The integral is $\int -\tan(\frac{\pi}{4}-x) dx$.
Let $u = \frac{\pi}{4}-x$. Then $du = -dx$, so $dx = -du$.
The integral becomes $\int -\tan(u) (-du) = \int \tan(u) du$.
$\int \tan(u) du = \log|\sec u| + C$.
Substituting back $u=\frac{\pi}{4}-x$:
$\log\left|\sec\left(\frac{\pi}{4}-x\right)\right| + C$.