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kaiz122

integrate(cos^3 x)/sin(x) dx

  • one year ago
  • one year ago

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  1. kaiz122
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    |dw:1339945551113:dw|

    • one year ago
  2. tishara429
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    bhyu

    • one year ago
  3. kaiz122
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    what?

    • one year ago
  4. SuckMyEsophagus
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    Let u = cos x du = -sin x dx then, "integral [u^3 -du]" or "- integral [u^3 du] " = -(u^4)/4 + C = -(cos^4 x)/4 + C

    • one year ago
  5. kaiz122
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    i have a question, sin x is in the denominator, how does it become the du then?

    • one year ago
  6. SuckMyEsophagus
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    Because everything is substituted and then flipped :)

    • one year ago
  7. kaiz122
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    i tried checking it by differentiation, then i got (cos^3x )(-sinx)

    • one year ago
  8. SuckMyEsophagus
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    Are you using your calculator?

    • one year ago
  9. kaiz122
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    no,

    • one year ago
  10. SuckMyEsophagus
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    You can integrate this directly, just notice that differentiating cos^4(x) gives you -4cos^3(x)sin(x) by the chain rule. Hence the integral of cos^3(x)sin(x) is -1/4 cos^4(x)

    • one year ago
  11. kaiz122
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    i still don't get it, how does sin x in the denominator becomes du? dx/sinx is not equal to sinx dx right.

    • one year ago
  12. SuckMyEsophagus
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    No their not the equal

    • one year ago
  13. kaiz122
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    @saifoo.khan can you please help me here

    • one year ago
  14. saifoo.khan
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    @dpaInc HELP!!!!!

    • one year ago
  15. dpaInc
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    |dw:1339951134590:dw|

    • one year ago
  16. kaiz122
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    thank you

    • one year ago
  17. AravindG
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    oh i was late :(

    • one year ago
  18. AravindG
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    use @AravindG if u have any integration qns in future

    • one year ago
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