mearing
Hey, I have a question about getting the correct solution for this practice problem:
\int 12cos^2(20x)dx
I started by applying the half-angle formula:
6\int (1+cos(40x))dx
Then i let u=40x so that \frac{1}{40}du=dx
=\frac{3}{20} \int (1+cos(u))du = \frac{3}{20} \int x+sin(u)du= \frac{3}{20} \int x+sin(40x)
= \frac{3}{20}x+\frac{3}{20}sin(40x)+C
The correct answer according to the Trig Tips link is :
6x+\frac{3}{20}sin(40x)+C
Is this the correct way to solve it?:
\int 12cos^2(20x)dx
Let u=40x so that \frac{1}{40}du=dx
=6\int (1+cos(40x))dx=6\int 1 dx +6\int cos(u))du = 6x +\frac{3}{20} \int (cos(u))du =6x+\frac{3}{20}sin(40x)+C
