integration of product of two dissimlar expressions

**flashylightsmeow** · March 30th 2011, 12:26 PM

I have to integrate by substitution:

$\int csc^2(3x+5)\frac{\pi}{4}dx<br />$ between 8/3 and 2

tried making (3x+5)=u to give:

$\int csc^2\frac{\pi u}{4}\frac{du}{3}<br />$

I'm not sure if this is correct though, and I don't know how to integrate trig functions when the variable has a coefficient. Please help.

**HallsofIvy** · March 30th 2011, 12:35 PM

So the $\frac{\pi}{4}$ is inside the csc? That is, it is $csc^2(\frac{\pi}{4}(3x+5))$ which is not quite what you wrote.

Once you have $\frac{1}{3}\int csc^2\frac{\pi u}{4}du$ make the further substitution $v= \frac{\pi u}{4}$ or, equivalently, make the original substitution $u= \frac{\pi}{4}(3x+ 5)$ rather than just u= 3x+ 5.

**flashylightsmeow** · March 30th 2011, 01:04 PM

Hi, the pi/4 is within the csc, yes. The answer is supposed to be 0 but i just cant seem to get that.

**HallsofIvy** · March 31st 2011, 05:38 AM

With the substitution $u= \frac{\pi}{4}(3x+ 5)$ , $du= \frac{3\pi}{4}dx$ so $dx= \frac{4}{3\pi}du$ . When x= 8/3, $u= \frac{\pi}{4}(8+ 5)= \frac{13\pi}{4}$ and when x= 2, $u= \frac{\pi}{4}(6+ 5)= \frac{11\pi}{4}$ . The integral becomes
$\frac{4}{3\pi}\int_{11\pi/4}^{13\pi/4} csc^2(u)du$

Now, the anti-derivative of $csc^2(u)$ is $- cot(u)$ . Evaluated at $\frac{13\pi}{4}$ that is 1 and at $\frac{11\pi}{4}$ it is -1. The value of the integral is NOT 0, it is $\frac{8}{3\pi}$ .

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