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∫(1/sinx)dx

Satyam Rout , 9 Years ago
Grade 12
anser 6 Answers
ranadeeshwar

Last Activity: 9 Years ago

1/2cos2x is thebest answer

Athiyaman

Last Activity: 9 Years ago

1/sinx can be written as cosecx and the integral of cosecx is log |cosecx – cotx|.
Therefore,
\int \frac{1}{sin(x)}dx = \int cosec(x)dx = \ln | cosecx - cotx| +c

Nikhil Upadhyay

Last Activity: 9 Years ago

\int 1/\sin x.dx=\int cosec x.dx=\left ( 1/a \right )\left [ log\left | \\cosec x-\cot x \right | \right ]+C

SREEKANTH

Last Activity: 8 Years ago

1/sin(x)=cosec (x)   the integration of cosec(x) is  ln|cosec x-cot x|+c  where cis an  arbitrary constant

kalyan

Last Activity: 8 Years ago

1/sinx =cosec x
or  integration of 1/x = ln |x|+c
where c is arbitari constant so by using this formula
integration of i/sinx is ln |sinx +c|

kalyan

Last Activity: 8 Years ago

sorry made a mistake currect answer is
1/sinx =cosec x
integration of cosec(x) is
ln|cosec x-cot x|+c
  where c is arbitari constant....

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