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## Zaira08 integrate integral sec^2 3 xdx /1+4tan3x 2 months ago 2 months ago

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1. goformit100

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2. myininaya

Try substituting the bottom

3. SithsAndGiggles

$\int\frac{\sec^23x}{1+4\tan3x}~dx$ Let $$u=1+4\tan 3x$$, so $$\dfrac{1}{12}du=\sec^23x~dx$$: $\frac{1}{12}\int\frac{du}{u}$