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 ?

#Calculus

Easy Math Editor

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Comments

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 - 4 years, 6 months ago

Yes, I approached the same way.

Kishore S. Shenoy - 4 years, 6 months ago

@Mark Hennings @Kishore S Shenoy @Calvin Lin

A Former Brilliant Member - 4 years, 6 months ago

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

Calvin Lin Staff - 4 years, 6 months ago

What is the ANS?????

Pathak Hardik - 4 years, 6 months ago

Mark Hennings has given it.

Kishore S. Shenoy - 4 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$