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Integral

Find \( \displaystyle \int \dfrac{dx}{4\sqrt2 \sin(3x) + 2\cos(3x)} \) it is easy but my answer is in different form from answer given , what do you get in terms of tanx and logs ? and by using substitutions ?

Note by Brilliant Member
11 months, 1 week ago

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Write \[ 4\sqrt{2}\sin3x + 2\cos3x \; = \; 6\cos(3x-\alpha) \hspace{2cm} \alpha = \cos^{-1}\tfrac13 \] and the integral becomes \[ \int \tfrac16\sec(3x-\alpha)\,dx \; = \; \tfrac{1}{18}\ln\big|\sec(3x-\alpha)+\tan(3x-\alpha)\big| + c \]

Mark Hennings - 11 months, 1 week ago

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Yes, I approached the same way.

Kishore S Shenoy - 11 months, 1 week ago

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@Mark Hennings @Kishore S Shenoy @Calvin Lin

Brilliant Member - 11 months, 1 week ago

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  1. Can you state the solution?
  2. Did you remember the "+C" which means that \( \sin^2 + C = - \cos^2 +C \)?

Calvin Lin Staff - 11 months, 1 week ago

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What is the ANS?????

Pathak Hardik - 11 months, 1 week ago

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Mark Hennings has given it.

Kishore S Shenoy - 11 months, 1 week ago

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