1. ## Integration

integrate sin3x/cos3x+sin3x from -pi by two to +

2. Originally Posted by sareeta
integrate sin3x/cos3x+sin3x from -pi by two to +
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(3x)}{\cos(3x)}+ \sin(3x)dx$

Rewrite this as $\int\tan(3x)+\sin(3x)dx$

we must have the derivative of the quantity out side before we can do anythign

so we have $\frac{1}{3}\int{3\tan(3x)}dx+\frac{1}{3}\int{3\sin (3x)}dx$

then by standard integration techniques we get

$\int\tan(3x)+\sin(3x)dx=\frac{-1}{3}\ln|\cos(3x)|-\frac{1}{3}\cos(3x)$

now just evaluate this from -pi/2 to pi/2

3. Originally Posted by sareeta
integrate sin3x/cos3x+sin3x from -pi by two to +
First I need to give thanks to the inte-killer for this trick

$\int\frac{\sin(3x)}{\cos(3x)+\sin(3x)}dx=\frac{1}{ 2}\int\frac{2\sin(3x)}{\cos(3x)+\sin(3x)}dx=$

$\frac{1}{2}\int\frac{2\sin(3x)+\cos(3x)-\cos(3x)}{\cos(3x)+\sin(3x)}dx=$

$\frac{1}{2}\int\frac{\sin(3x)+\cos(3x)}{\sin(3x)+\ cos(3x)}dx+\frac{1}{2}\int \frac{\sin(3x)-\cos(3x)}{\sin(3x)+\cos(3x)}dx=$

$\frac{1}{2}\int dx -\frac{1}{2}\int \frac{-\sin(3x)+\cos(3x)}{\sin(3x)+\cos(3x)}dx$

making a u sub on gives this as an antiderivative

$\frac{1}{2}x-\frac{1}{6}\ln|\sin(3x)+\cos(3x)|$

I'm not sure what you mean by your limits of integration.

Good luck.

4. Originally Posted by sareeta
integrate sin3x/cos3x+sin3x from -pi by two to +
There is a slicker solution since its a definite integral

Let $f(x) = \frac{\sin(x)}{\cos(x)+\sin(x)}$

So we want $I = \int_{-\frac{\pi}2}^{\frac{\pi}2} f(3x) \, dx$

Lets u-sub $u = 3x$ and break the integral into two pieces,

$I_1 = \frac13 \int_{-\frac{3\pi}2}^{0} f(u) \, du$ and $I_2 = \frac13 \int_{0}^{\frac{3\pi}2} f(u) \, du$

$I_1 = \frac13 \int_{-\frac{3\pi}2}^{0} f(u) \, du = \frac13 \int_{-\frac{3\pi}2}^{0} f\left(-\frac{3\pi}2 - u\right) \, du =
\frac13 \int_{-\frac{3\pi}2}^{0} (1 - f(u)) \, du$

$\Rightarrow 2I_1 = \frac13 \int_{-\frac{3\pi}2}^{0} \, du = \frac13 .\frac{3\pi}2
\Rightarrow I_1 = \frac{\pi}4$

Clearly $I_2 - I_1 = 0 \Rightarrow I_1 = I_2 = \frac{\pi}4$
So $I = I_1 + I_2 = \frac{\pi}2$

5. Hello,

Originally Posted by TheEmptySet
First I need to give thanks to the inte-killer for this trick

$\int\frac{\sin(3x)}{\cos(3x)+\sin(3x)}dx=\frac{1}{ 2}\int\frac{2\sin(3x)}{\cos(3x)+\sin(3x)}dx=$

$\frac{1}{2}\int\frac{2\sin(3x)+\cos(3x)-\cos(3x)}{\cos(3x)+\sin(3x)}dx=$

$\frac{1}{2}\int\frac{\sin(3x)+\cos(3x)}{\sin(3x)+\ cos(3x)}dx+\frac{1}{2}\int \frac{\sin(3x)-\cos(3x)}{\sin(3x)+\cos(3x)}dx=$

$\frac{1}{2}\int dx -\frac{1}{2}\int \frac{-\sin(3x)+\cos(3x)}{\sin(3x)+\cos(3x)}dx$

making a u sub on gives this as an antiderivative

$\frac{1}{2}x-\frac{1}{\color{red}6}\ln|\sin(3x)+\cos(3x)|$

I'm not sure what you mean by your limits of integration.

Good luck.
There's a 3 missing (red)