Thread: Need help with an integral View Single Post
 2011-04-10, 19:56 #4 Mathew     Nov 2009 2·52·7 Posts $\int_{b}^aA\sin\left(\frac{\pi(t-6)}{12}\right)+B dt$ what I understood from Baz's post $\int_{a}^bA\sin\left(\frac{t\pi-6}{12}\right)+B dt$ what Christenson wrote Let us look at Christenson's equation without the points $\int_{}A\sin\left(\frac{t\pi-6}{12}\right)+Bdt$ Break this into two integrals $\int_{}A\sin\left(\frac{t\pi-6}{12}\right)dt+\int_{}Bdt$ Let us do the harder integral first $\int_{}A\sin\left(\frac{t\pi-6}{12}\right)dt$ A is a constant and can be placed outside of the integral $A\int_{}\sin\left(\frac{t\pi-6}{12}\right)dt$ Next is an u-substitution let $u=\frac{t\pi-6}{12}$ then $du=\frac{\pi}{12}dt$ Therefore $dt=\frac{12}{\pi}du$ Substitute the values $A\int_{}\frac{12}{\pi}\sin\left(u\right)du$ Again since $\frac{12}{\pi}$ is a constant move it out of the integral Now you have $A\frac{12}{\pi}\int_{}\sin\left(u\right)du$ Which is $-A\frac{12}{\pi}\cos\left(u\right)$ replace the u for what was substituted $-A\frac{12}{\pi}\cos\left(\frac{t\pi-6}{12}\right)$ Now do the second integral $\int_{}Bdt =Bt$ The whole equation is $-A\frac{12}{\pi}\cos\left(\frac{t\pi-6}{12}\right)+Bt$ Evaluating from a to b i.e. $\int_{a}^b$ it would be the following $\left(-A\frac{12}{\pi}\cos\left(\frac{b\pi-6}{12}\right)+B*b \right)-\left(-A\frac{12}{\pi}\cos\left(\frac{a\pi-6}{12}\right)+B*a\right)$ Entering in all the values =6.25624