reduction method to solve the integral sin^2(17x)dx

xomandi

New member
Joined
Oct 1, 2006
Messages
31
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???
 
Hello, xomandi!

Use the reduction formula to solve: .\(\displaystyle \L\int\)sin2(17x)dx\displaystyle \sin^2(17x)\,dx

What "reduction formula" are they referring to?

This Double-Angle Identity is appropriate: sin2θ=1cos2θ2\displaystyle \:\sin^2\theta \:=\:\frac{1\,-\,\cos2\theta}{2}


Then we have: \(\displaystyle \L\:\frac{1}{2}\int \left[1\,-\,\cos(34x)\right]\,dx\)

 
Top