Calculus I Review Evaluate the Integrals 1. ∫ xsin(x2)dx Solution: u

Calculus I Review Evaluate the Integrals 1. ∫ xsin(x2)dx Solution: u

Calculus I Review
Evaluate the Integrals
Z
1.
x sin(x2 )dx
Solution: u = x2 , du = 2xdx ⇒
Z
Z
1
1
1
x sin(x2 )dx =
sin udu = − cos u + C = − cos x2 + C
2
2
2
√
sin x
√ dx
2.
x
√
Solution: u = x, du =
Z
Z
Z
3.
1 √1
2 x dx
⇒
√
Z
√
sin x
√ dx = 2 sin udu = −2 cos u + C = −2 cos x + C
x
tan2 θ sec2 θdθ
Solution: u = tan θ, du = sec2 θdθ ⇒
Z
Z
1
1
2
2
tan θ sec θdθ = u2 du = u3 + C = tan3 θ + C
3
3
Z
4.
ecos t sin tdt
Solution: u = cos t, du = − sin t ⇒
Z
Z
ecos t sin tdt = − eu du = −eu + C = −ecos t + C
Z
2
1
ex
5.
dx
2
1 x
Solution: u = x1 , du = − x12 ⇒
Z
1
2
1
ex
dx = −
x2
Z
1
2
1
eu du = e − e 2
1
Z
6.
sin t sec(cos t) tan(cos t)dt
Solution: u = cos t, du = − sin tdt ⇒
Z
Z
sin t sec(cos t) tan(cos t)dt = sec u tan udu = − sec u+C = − sec(cos t)+C
Z
1
dx.
x
Solution:
ln x + C
Z
sin x
8.
dx.
1 + cos2 x
Solution: u = cos x, du = − sin xdx ⇒
Z
Z
sin x
du
dx = −
= − tan−1 u + C = − tan−1 (cos x) + C
2
1 + cos x
1 + u2
Z
√
9.
ex 1 + ex dx.
7.
Solution: u = 1 + ex , du = ex dx ⇒
Z
Z
√
√
3
2 3
2
x
x
e 1 + e dx =
udu = u 2 = (1 + ex ) 2
3
3