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∫(1/sinx)dx

∫(1/sinx)dx

Grade:12

6 Answers

ranadeeshwar
19 Points
5 years ago
1/2cos2x is thebest answer
Athiyaman
57 Points
5 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
171 Points
5 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
85 Points
4 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
37 Points
4 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
37 Points
4 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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