Answer to Question #87992 in Calculus for Joshua

Question

1.Find the integral with respect to x ∫cos x sin xdx
a.sin^2x
-------- +c
2
b.sin2x+c
c.cos^2x
----- +c
2
d.sinx

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

1. Use the substitution rule. Let

$$\sin x=u$$

This implies that

$$\cos xdx=d\left( \sin x \right)=du$$

Then

$$\int \cos x\sin x\,dx=\int \underbrace{\sin x}_{u}\overbrace{\cos x\,dx}^{du}=\int u\,du=\frac{{{u}^{2}}}{2}+c=\frac{{{\sin }^{2}}x}{2}+c$$

Answer: a.

 

2. Use the substitution rule. Let

$$\left( 6x+4 \right)=u$$

Find the differential dx

$$du=d\left( 6x+4 \right)=6dx\,\,=>\,\,dx=\frac{1}{6}du$$

Then

$$\int \cos \left( 6x+4 \right)dx=\int \cos u\cdot \frac{1}{6}du$$

$$=\frac{1}{6}\int \cos udu=\frac{1}{6}\sin u+c=\frac{1}{6}\sin \left( 6x+4 \right)+c$$

Answer: a.