The value of 2∫sin⁡xdxsin⁡x−π4 is equal to

The value of $\sqrt{2}\int \frac{\mathrm{sin}\mathrm{xdx}}{\mathrm{sin}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)}$ is equal to

1. A

$\mathrm{x}-\mathrm{log}\left|\mathrm{cos}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)\right|+\mathrm{C}$

2. B

$\mathrm{x}+\mathrm{log}\left|\mathrm{cos}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)\right|+\mathrm{C}$

3. C

$\mathrm{x}-\mathrm{log}\left|\mathrm{sin}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)\right|+\mathrm{C}$

4. D

$\mathrm{x}+\mathrm{log}\left|\mathrm{sin}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)\right|+\mathrm{C}$

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Solution:

Let $\mathrm{I}=\sqrt{2}\int \frac{\mathrm{sin}\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)}\mathrm{dx}$

Put, $\mathrm{x}=\frac{\mathrm{\pi }}{4}=\mathrm{t}⇒\mathrm{dx}=\mathrm{dt}$

$\begin{array}{l}\therefore \mathrm{I}=\sqrt{2}\int \frac{\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}+\mathrm{t}\right)\mathrm{dt}}{\mathrm{sin}\mathrm{t}}\\ =\sqrt{2}\int \left[\frac{1}{\sqrt{2}}\mathrm{cot}\mathrm{t}+\frac{1}{\sqrt{2}}\right]\mathrm{dt}\\ =\mathrm{log}|\mathrm{sin}\mathrm{t}|+\mathrm{t}+\mathrm{C}\\ =\mathrm{x}+\mathrm{log}\left|\mathrm{sin}\left(\mathrm{x}-\frac{\mathrm{\pi }}{4}\right)\right|+\mathrm{C}\end{array}$

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