Integral Calculus Question 14
Question: $ \int{{{\cos }^{-3/7}}}x{{\sin }^{-11/7}}xdx= $
[Kerala (Engg.) 2005]
Options:
A) $ \log |{{\sin }^{4/7}}x|+c $
B) $ \frac{4}{7}{{\tan }^{4/7}}x+c $
C) $ \frac{-7}{4}{{\tan }^{-4/7}}x+c $
D) $ \log |{{\cos }^{3/7}}x|+c $
E) $ \frac{7}{4}{{\tan }^{-4/7}}x+C $
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Answer:
Correct Answer: C
Solution:
$ m+n=-\frac{3}{7}+( \frac{-11}{7} )=-2 $ (?ve integer) $ I=\int{{{\cos }^{-3/7}}x( {{\sin }^{(-2+3/7)}}x )dx}=\int{{{\cos }^{-3/7}}}x,{{\sin }^{-2}}x,{{\sin }^{3/7}}xdx $ $ =\int{\frac{\cos ec^{2}x}{( \frac{{{\cos }^{3/7}}x}{{{\sin }^{3/7}}x} )}}dx=\int{\frac{\cos ec^{2}x,dx}{{{\cot }^{3/7}}x}} $ Put $ \cot x=t $ Þ $ -\cos ec^{2}xdx=dt $ $ I=-\int{\frac{dt}{{t^{3/7}}}} $ $ =-\frac{{t^{-\frac{3}{7}+1}}}{-\frac{3}{7}+1}+c $ $ =-\frac{7}{4}{t^{4/7}}+c $ $ =-\frac{7}{4}{{\cot }^{4/7}}x+c $ $ =-\frac{7}{4}{{\tan }^{-4/7}}x+c $ .
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