TS EAMCET · Maths · Indefinite Integration
If \(\frac{5 \pi}{4} < x < \frac{7 \pi}{4}\), then \(\int \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x=\)
- A \(-\sec ^2\left(\frac{\pi}{4}-x\right)+\mathrm{c}\)
- B \(-\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)
- C \(\sec ^2\left(\frac{\pi}{4}-x\right)+c\)
- D \(\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)
Answer & Solution
Correct Answer
(D) \(\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)
Step-by-step Solution
Detailed explanation
\( \int \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x = \int \sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} d x \) \( = \int \left|\frac{\cos x - \sin x}{\cos x + \sin x}\right| d x = \int \left|\frac{1 - \tan x}{1 + \tan x}\right| d x \)…
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