X xomandi New member Joined Oct 1, 2006 Messages 31 Nov 13, 2007 #1 Use the reduction formula to solve the integral sin^2(17x)dx so would it be then... -1/2 cos (17x) sin(17x) + 1/2 int xdx???
Use the reduction formula to solve the integral sin^2(17x)dx so would it be then... -1/2 cos (17x) sin(17x) + 1/2 int xdx???
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 13, 2007 #2 Hello, xomandi! Use the reduction formula to solve: .\(\displaystyle \L\int\)\(\displaystyle \sin^2(17x)\,dx\) Click to expand... What "reduction formula" are they referring to? This Double-Angle Identity is appropriate: \(\displaystyle \:\sin^2\theta \:=\:\frac{1\,-\,\cos2\theta}{2}\) Then we have: \(\displaystyle \L\:\frac{1}{2}\int \left[1\,-\,\cos(34x)\right]\,dx\)
Hello, xomandi! Use the reduction formula to solve: .\(\displaystyle \L\int\)\(\displaystyle \sin^2(17x)\,dx\) Click to expand... What "reduction formula" are they referring to? This Double-Angle Identity is appropriate: \(\displaystyle \:\sin^2\theta \:=\:\frac{1\,-\,\cos2\theta}{2}\) Then we have: \(\displaystyle \L\:\frac{1}{2}\int \left[1\,-\,\cos(34x)\right]\,dx\)
