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Zaira08

  • one year ago

integrate integral sec^2 3 xdx /1+4tan3x

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  1. goformit100
    • one year ago
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  2. myininaya
    • one year ago
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    Try substituting the bottom

  3. SithsAndGiggles
    • one year ago
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    \[\int\frac{\sec^23x}{1+4\tan3x}~dx\] Let \(u=1+4\tan 3x\), so \(\dfrac{1}{12}du=\sec^23x~dx\): \[\frac{1}{12}\int\frac{du}{u}\]

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