Question

$ \int{\frac{(\sin \theta +\cos \theta )}{\sqrt{\sin 2\theta }}}d\theta $ is equal to:

• A. $ \log |\cos \theta -\sin \theta +\sqrt{\sin 2\theta }|+c $
• B. $ \log |\sin \theta -\cos \theta +\sqrt{\sin 2\theta }|+c $
• C. $ {{\sin }^{-1}}(\sin \theta -\cos \theta )+c $
• D. $ {{\sin }^{-1}}(\sin \theta +\cos \theta )+c $
• E. $ {{\sin }^{-1}}(\cos \theta -\sin \theta )+c $

Answer 1

Answer: (C)

Let $ I=\int{\frac{\sin \theta +\cos \theta }{\sqrt{1+\sin 2\theta -1}}d\theta } $
$ =\int{\frac{\sin \theta +\cos \theta }{\sqrt{1-{{(\sin \theta -\cos \theta )}^{2}}}}d\theta } $
Let $ \sin \theta -\cos \theta =t $
$ \Rightarrow $ $ (\cos \theta +\sin \theta )d\theta =dt $
$ \therefore $ $ I=\int{\frac{1}{\sqrt{1-{{t}^{2}}}}}dt={{\sin }^{-1}}(\sin \theta -\cos \theta )+c $