Integral Calculus Question 12
Question: $ \int{\frac{(\sin \theta +\cos \theta )}{\sqrt{\sin 2\theta }}}d\theta = $
[Kerala (Engg.) 2005]
Options:
A) $ \log | \cos \theta -\sin \theta +\sqrt{\sin 2\theta } | $
B) $ \log | \sin \theta -\cos \theta )+\sqrt{\sin 2\theta } | $
C) $ {{\sin }^{-1}}(\sin \theta -\cos \theta )+c $
D) $ {{\sin }^{-1}}(\sin \theta +\cos \theta )+c $
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Answer:
Correct Answer: C
Solution:
Let $ I=\int{\frac{\sin \theta +\cos \theta }{\sqrt{2\sin \theta \cos \theta }}d\theta } $ \ $ I=\int{\frac{\sin \theta +\cos \theta }{\sqrt{1-(1-2\sin \theta \cos \theta )}}d\theta } $ $ =\int{\frac{(\sin \theta +\cos \theta )d\theta }{\sqrt{1-({{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta )}}} $ $ =\int{\frac{\sin \theta +\cos \theta }{\sqrt{1-{{(\sin \theta -\cos \theta )}^{2}}}}d\theta } $ Let $ (\sin \theta -\cos \theta )=t $
Þ $ (\cos \theta +\sin \theta )d\theta =dt $ \ $ I=\int{\frac{dt}{\sqrt{1-t^{2}}}={{\sin }^{-1}}(t)+c}={{\sin }^{-1}}(\sin \theta -\cos \theta )+c $ .
BETA
