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kaiz122

  • 2 years ago

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

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  1. kaiz122
    • 2 years ago
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    |dw:1339945551113:dw|

  2. tishara429
    • 2 years ago
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    bhyu

  3. kaiz122
    • 2 years ago
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    what?

  4. SuckMyEsophagus
    • 2 years ago
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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

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

  6. SuckMyEsophagus
    • 2 years ago
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    Because everything is substituted and then flipped :)

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

  8. SuckMyEsophagus
    • 2 years ago
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    Are you using your calculator?

  9. kaiz122
    • 2 years ago
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    no,

  10. SuckMyEsophagus
    • 2 years ago
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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)

  11. kaiz122
    • 2 years ago
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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.

  12. SuckMyEsophagus
    • 2 years ago
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    No their not the equal

  13. kaiz122
    • 2 years ago
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    @saifoo.khan can you please help me here

  14. saifoo.khan
    • 2 years ago
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    @dpaInc HELP!!!!!

  15. dpaInc
    • 2 years ago
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    |dw:1339951134590:dw|

  16. kaiz122
    • 2 years ago
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    thank you

  17. AravindG
    • 2 years ago
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    oh i was late :(

  18. AravindG
    • 2 years ago
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    use @AravindG if u have any integration qns in future

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