Answer to Question #88004 in Math for Joshua

Question

2.Evaluate ∫cos(6x+4)dx
a.sin(6x+4)
--------- +c
6

b.cos(6x+4)
--------- +c
6
c.tan(6x+4)
--------- +c
6
d.sec(6x+4)
--------- +c
6

Answer 1

Solution. let y=6x+4 get dy=6dx. We make a substitution and get

$$\int cos(6x+4)dx=\frac {1} {6} \int cos(y)dy$$

Using a table of integrals

$$\int cos(x)dx=sin(x)+C$$

where C is constant. Therefore

$$\frac {1} {6} \int cos(y)dy=\frac {1} {6} sin(y)+C$$

Returning to the substitution we get

$$\frac {1} {6} sin(y)+C=\frac {1} {6} sin(6x+4)+C$$

where C is constant.

Answer. a.

$$\frac {1} {6} sin(6x+4) + C$$

where C is constant.