The value of + 1- 2x dx is equal to: - Vidyadip

MCQ

The value of $\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-\sin2\text{x}}}\text{ dx}$ is equal to:

A
$\sqrt{\sin2\text{x}}+\text{C}$
B
$\sqrt{\cos2\text{x}}+\text{C}$
C
$\pm(\sin\text{x}-\cos\text{x})+\text{C}$
✓
$\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$

✓

Answer

Correct option: D.

$\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$

Let $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-\sin2\text{x}}}\text{ dx}$
$=\int\frac{(\sin\text{x}+\cos\text{x})\text{dx}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}}}$
$=\int\frac{(\sin\text{x}+\cos\text{x})}{\sqrt{(\sin\text{x}-\cos\text{x})^2}}$
$=\int\frac{(\sin\text{x}+\cos\text{x})\text{dx}}{|\sin\text{x}-\cos\text{x}|}$
$=\pm\int\Big(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$
$(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\pm\int\frac{\text{dt}}{\text{t}}$
$=\ln|\text{t}|+\text{C}$
$=\pm\ln(\sin\text{x}-\cos\text{x})+\text{C}$ $(\because\text{t}=\sin\text{x}-\cos\text{x})$

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