Question

What am I doing wrong with this integral problem?

Answer 1

..............

Answer 2

The problem is \[\int\limits (\sin(3x))/(16+\cos^{2}(3x))\] so here is what I did:
\[1/3 \int\limits (1/(16+\cos^{2}(x)) * \sin(x)\]
\[1/3 \int\limits (1/4)(\arctan(\cos(x)/4) * -\cos(x)\]
\[1/12 (\arctan(\cos(x)/4) * -\cos(x)\]
but the answer my teacher has is \[-1/12 (\arctan(\cos(x)/4)\]

Answer 3

\[\int\limits_{}^{}\frac{\sin(3x)}{16+\cos^2(3x)} dx\]
\[\text{ Let } 4u=\cos(3x) => 4 du=-3 \sin(3x) dx\]
So we have
\[\int\limits_{}^{}\frac{-4}{3} \cdot \frac{du}{16+16u^2}\]

Answer 4

\[\frac{-4}{3} \cdot \frac{1}{16} \int\limits_{}^{}\frac{1}{1+u^2} du\]

Answer 5

your first line needs another letter like u = 3x for clarity since 3sin(x) is not sin(3x) and so on.

Similarly on the second line it seems like you're trying to integrate something that is already the answer to the integral, and so you're thinking you need an extra chain rule term.

Instead, just let 4u = cos(3x) explicitly as per myininaya and then let tan(theta) = u. It's never efficient to skip u-substitution steps...

Answer 6

\[-\frac{1}{3} \cdot \frac{1}{4} \arctan(u)+C=\frac{-1}{12}\arctan(\frac{\cos(3x)}{4})+C\]

Answer 7

yes we can use that other substitution broken was talking about after the 1st i made

Answer 8

(even your teacher answer is wrong, due to 3x <-> x sloppiness)

Answer 9

so instead we can use one substitution instead of two
we have
\[\int\limits\limits_{}^{}\frac{\sin(3x)}{16+\cos^2(3x)} dx \]
\[\cos(3x)=4 \tan(3u) =>-3 \sin(3x) dx=12 \sec^2(3u) du\]
\[\int\limits_{}^{}\frac{-4 \sec^2(3u)}{16+16 \tan^2(3u)} du=\int\limits_{}^{}\frac{-4 \sec^2(3u)}{16(1+\tan^2(3u))} du\]
\[=\frac{-1}{4} \int\limits_{}^{}\frac{\sec^2(3u)}{\sec^2(3u)} du=\frac{-1}{4}\int\limits_{}^{}1 du =-\frac{1}{4} u+C\]
then we can solve our equation \[\cos(3x)=4 \tan(3u)\] for u
\[\frac{1}{4} \cos(3x)=\tan(3u) => \arctan(\frac{\cos(3x)}{4})=3u=>\frac{1}{3} \arctan(\frac{\cos(3x)}{4})=u\]

Answer 10

=>\[\frac{-1}{4}u+C=\frac{-1}{4} \cdot \frac{1}{3} \arctan(\frac{\cos(3x)}{4})+C\]

Answer 11

\[={1 \over 12}arctan(4sec(3x))+C\]

Answer 12

Assuming both results are correct and noticing the trig identity\[4\sec(3x) = \frac{1}{\frac{\cos(3x)}{4}},\]We can set @myininaya equal to @UnkleRhaukus to obtain the trig identity \[\arctan(\theta) = -\arctan(\frac{1}{\theta}) + C.\]Presumably \[C=\frac{\pi}{2}.\]