252 STEP 4. Review the Knowledge You Need to Score High
- Letu= 1 +cosx;du=−sinxdxor
−du=sinxdx.
∫
sinx
√
1 +cosx
dx=∫
− 1
√
u
(du)=−
∫
1
u^1 /^2
du=−
∫
u−^1 /^2 du=−
u^1 /^2
1 / 2+C
=− 2 u^1 /^2 +C
=−2(1+cosx)^1 /^2 +C
∫π0√sinx
1 +cosxdx=−2(1+cosx)^1 /^2]π
0=− 2[
(1+cosπ)^1 /^2
−(1+cos 0)^1 /^2]=− 2[
0 − 21 /^2]
= 2√
2- Letu= 4 x;du= 4 dxor
du
4
=dx.
∫
g(4x)dx=
∫
g(u)
du
4=
1
4
∫
g(u)du=
1
4
f(u)+C=
1
4
f(4x)+C∫ 21g(4x)dx=1
4
f(4x)]^21=
1
4
f(4(2))−1
4
(f(4(1))=
1
4
f(8)−1
4
f(4)8.
∫ln 3ln 210 exdx= 10 ex]ln 3
ln 2= 10[(
eln 3)
−(
eln 2)]=10(3−2)= 10
9.Letu=t+3;du=dt.
∫
1
t+ 3
dt=∫
1
u
du=ln|u|+C=ln∣∣
t+ 3∣∣
+C
∫e 2e1
t+ 3
dt=ln∣∣
t+ 3∣∣]e 2
e=ln(e^2 +3)−ln(e+3)=ln(
e^2 + 3
e+ 3)- f′(x)=tan^2 x;
f′(
π
6)
=tan^2(
π
6)
=(
1
√
3) 2
=1
3
- Letu=x^2 ;du= 2 xdxor
du
2
=xdx.
∫
4 xex
2
dx= 4
∫
eu(
du
2)= 2
∫
eudu= 2 eu+c= 2 ex^2 +C
∫ 1− 14 xex^2 dx= 2 ex^2] 1
− 1= 2[
e(1)
2
−e(−1)
2 ]
=2(e−e)= 0Note thatf(x)= 4 xex^2 is an odd function.
Thus,∫a−af(x)dx=0.12.
∫π−π(
cosx−x^2)
dx=sinx−
x^3
3]π−π=(
sinπ−
π^3
3)−
(
sin(−π)−
(−π)^3
3)=−
π^3
3−
(
0 −
−π^3
3)=−
2 π^3
3