Integration question

Hey,
Looking at this integrtion question

The temperature (in degrees cenigrade) measured at a weather station on 21st Sept is given by

T(t) = 15 - 10cos^2(pi*t/24) - t/10

0 < t < 24

where t is the time measured in hours. Find average temperature over the entire day, and average during the hours of daylight 6<t<18

It's just integrating the expression I can't do, the worst part is I have my working from when I have done it before but I haven't copied it all down and now I can't see how I integrated that expression last time to get the answer :|

Thanks for any help

Use double angle formulae for $cos^2$

$cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1$ and rearrangle for cos squared.

and $x \mapsto \frac{\pi}{24}\cdot t$

Ah yeah so that gives me

T(t) = 15 - 5 + 5cos(pi*t/12) - t/10
= 10 + + 5cos(pi*t/12) - t/10

thanks!

I'm just a little unsure how to do the, + 5cos(pi*t/12)? Sorry my brain is failing as i'm sitting looking at my previous correct answer for this!

Reverse chain rule. Cos goes to sine and you divide the 5 by the coefficient of t; in this case pi/12

Reverse chain rule. Cos goes to sine and you divide the 5 by the coefficient of t; in this case pi/12

The chain rule in general doesn't work in reverse: the method that you have described only works when the derivative of the inner function is a constant, or in other words, the inner function is linear. What you are really doing is making a substitution and skipping out the steps that you can do in your head.