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Trig integral practice question

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 $\frac{1}{40}du=dx$

$=6\int (1+cos(40x))dx=6\int 1 dx +6\int cos(u))du = 6x +\frac{3}{20} \int (cos(u))du =6x+\frac{3}{20}sin(40x)+C$

mark

This looks perfect!!