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Practice  Integrals         Math  252,    Oregon  State  University     The  first  30  integrals  do  not  require  the  use  of  either  trigonometric  substitution  or   partial  fractions.     1.   ∫e 5− 4 x dx   2.   ∫ sin(π z) 4.   ∫ xe 6.   ∫ dz     3.   ∫ cos (3θ ) 2 dθ   −2 x dx     5.   ∫t 3 ln t dt   4x 9 − 4x 2 dx     7.   ∫ dx   8.   et − e−t ∫ et + e−t dt   dα   10.   ∫ 1+ x 12.   ∫ sin 14.   ∫ 9 − 4x 2   9.   cos α ∫ 1 + sin 2 ∫ x tan x dx   α x 4 dx     11.   −1 3 ϕ dϕ     13.   ∫ sec 4 t dt   tan ( z ) dz   z   15.   ∫5 2x dx   16.       Page  99   ∫e 3x ( ) sec e3x dx   1 3 18.   ∫ y ( 2 − y ) 2 dy   3x ∫ e sin(2x) dx   20.   t 2 + 4t + 3 ∫ t + 1 dt   ∫x e dx   22.   ∫ tan (3θ ) dα ∫ sec(2α )   24.   −( x −1) dx   ∫ (x − 1)e 25.   ∫ tan θ sec 2 θ dθ   26.   ∫ (w 27.   ∫ ln 7 dx   28.   ∫ sec(2θ ) 30.   ∫ tan 17.   ∫ 1+ dx   x   19.     21.   2 5x dθ   2   23.   2   2 2 ) + 2w + 1 w + 1 dw   dθ     29.   ∫ ln(5x) dx   5 θ sec 3 θ dθ     The  next  20  integrals  may  require  either  trigonometric  substitution  or  partial   fractions.     31.   x+3 ∫ x 2 + 2x + 5 dx   32.   t3 ∫ t 2 + 1 dt   1+ x dx   1− x 34.   ∫z+   33.   ∫     Page  100   dz   z
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𝑑𝑢

1

1

𝟏

1. ∫ 𝑒 5−4𝑥 𝑑𝑥 = |𝑢 = 5 − 4𝑥, 𝑑𝑢 = −4𝑑𝑥| = ∫ 𝑒 𝑢 −4 = − 4 ∫ 𝑒 𝑢 𝑑𝑢 = − 4 𝑒 𝑢 + 𝐶 = − 𝟒 𝒆𝟓−𝟒𝒙 + 𝐶.

2. ∫ sin(𝜋𝑧) 𝑑𝑧 = |𝑢 = 𝜋𝑧, 𝑑𝑢 = 𝜋𝑑𝑧| = ∫ sin 𝑢

3. ∫ cos 2 (3𝜃) 𝑑𝜃 = ∫

1+cos(6𝜃)
2

𝟏

𝑑𝑢
𝜋

1

1

𝟏

= 𝜋 ∫ sin 𝑢 𝑑𝑢 = − 𝜋 cos 𝑢 + 𝐶 = − 𝝅 𝐜𝐨𝐬(𝝅𝒛) + 𝐶.

𝟏

𝑑𝜃 = 𝟐 𝜽 + 𝟏𝟐 𝐬𝐢𝐧(𝟔𝜽) + 𝐶.

1

1

1

𝟏

𝟏

4. ∫ 𝑥𝑒 −2𝑥 𝑑𝑥 = |𝑢 = 𝑥, 𝑑𝑢 = 𝑑𝑥, 𝑑𝑣 = 𝑒 −2𝑥 𝑑𝑥, 𝑣 = − 2 𝑒 −2𝑥 | = − 2 𝑥𝑒 −2𝑥 + 2 ∫ 𝑒 −2𝑥 𝑑𝑥 = − 𝟐 𝒙𝒆−𝟐𝒙 − 𝟒 𝒆−𝟐𝒙 + 𝐶.

5. ∫ 𝑡 3 ln 𝑡 𝑑𝑡 = |𝑢 = ln 𝑡 , 𝑑𝑢 =

𝑑𝑡
𝑡

1

4𝑥

6. ∫ √9−4𝑥 2 𝑑𝑥 = |𝑢 = 9 − 4𝑥 2 , 𝑑𝑢 = −8𝑥𝑑𝑥| = − ∫

𝑑𝑥

2

2

1

1

, 𝑑𝑣 = 𝑡 3 𝑑𝑡, 𝑣 = 4 𝑡 4 | = 4 𝑡 4 ln 𝑡 − 4 ∫ 𝑡 4

𝑡

𝟏

𝟏

= 𝟒 𝒕𝟒 𝐥𝐧 𝒕 − 𝟏𝟔 𝒕𝟒 + 𝐶.

= −√𝑢 + 𝐶 = −√𝟗 − 𝟒𝒙𝟐 + 𝐶.

√𝑢

3
𝑑𝑢
2

3

7. ∫ √9−4𝑥 2 = |𝑢 = 3 𝑥, 𝑑𝑢 = 3 𝑑𝑥, 𝑥 = 2 𝑢| = ∫

2𝑑𝑢

𝑑𝑡

1

2

√9−4(3𝑢)

𝑑𝑢

1

𝟏

𝟐

= 2 ∫ √1−𝑢2 = 2 arcsin 𝑢 + 𝐶 = 𝟐 𝐚𝐫𝐜𝐬𝐢𝐧 (𝟑 𝒙) + 𝐶.

2

𝑒 𝑡 −𝑒 −𝑡

sinh 𝑡

8. ∫ 𝑒 𝑡 +𝑒 −𝑡 𝑑𝑡 = ∫ cosh 𝑡 𝑑𝑡 = |𝑢 = cosh 𝑡 , 𝑑𝑢 = sinh 𝑡 𝑑𝑡| = ∫

cos 𝛼

𝑑𝑢
𝑢

= ln|𝑢| + 𝐶 = ln|cosh 𝑡| + 𝐶 = ln(cosh 𝑡) + 𝐶 = 𝐥𝐧(𝒆𝒕 + 𝒆−𝒕 ) + 𝐶2 .

𝑑𝑢

9. ∫ 1+sin2 𝛼 𝑑𝛼 = |𝑢 = sin 𝛼 , 𝑑𝑢 = cos 𝛼 𝑑𝛼| = ∫ 1+𝑢2 = arctan 𝑢 + 𝐶 = 𝐚𝐫𝐜𝐭𝐚𝐧(𝐬𝐢𝐧 𝜶) + 𝐶.

𝑥

1

𝑑𝑢

1

𝟏

2
𝟐)
10. ∫ 1+𝑥 4 𝑑𝑥 = |𝑢 = 𝑥 2 , 𝑑𝑢 = 2𝑥𝑑𝑥| = ∫ 1+𝑢
+ 𝐶.
2 = 2 arctan 𝑢 + 𝐶 = 𝟐 𝐚𝐫𝐜𝐭𝐚𝐧(𝒙

𝑑𝑥

1

1

𝑥2

1

1

1

1

11. ∫ 𝑥 arctan 𝑥 𝑑𝑥 = |𝑢 = arctan 𝑥 , 𝑑𝑢 = 1+𝑥 2 , 𝑑𝑣 = 𝑥𝑑𝑥, 𝑣 = 2 𝑥 2 | = 2 𝑥 2 arctan 𝑥 − 2 ∫ 1+𝑥 2 𝑑𝑥 = 2 𝑥 2 arctan 𝑥 − 2 ∫ (1 − 1+𝑥 2 ) 𝑑𝑥 =
1
1
𝟏
𝟏
= 𝑥 2 arctan 𝑥 − (𝑥 − arctan 𝑥) + 𝐶 = (𝒙𝟐 + 𝟏) 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 − 𝒙 + 𝑪.
2
2
𝟐
𝟐
12. ∫ sin3 𝜑 𝑑𝜑 = |𝑢 = cos 𝜑 , 𝑑𝑢 = − sin 𝜑 𝑑𝜑, sin2 𝜑 = 1 − cos2 𝜑 = 1 − 𝑢2 | =
1
𝟏
= − ∫(1 − 𝑢2 )𝑑𝑢 = −𝑢 + 𝑢3 + 𝐶 = − 𝐜𝐨𝐬 𝝋 + 𝐜𝐨𝐬𝟑 𝝋 + 𝐶.
3
𝟑

𝟏

13. ∫ sec 4 𝑡 𝑑𝑡 = ∫ sec 2 𝑡 (1 + tan2 𝑡)𝑑𝑡 = ∫ sec 2 𝑡 𝑑𝑡 + ∫ sec 2 𝑡 tan2 𝑡 𝑑𝑡 = tan 𝑡 + ∫ tan2 𝑡 𝑑(tan 𝑡) = 𝐭𝐚𝐧 𝒕 + 𝟑 𝐭𝐚𝐧𝟑 𝒕 + 𝐶.

14. ∫

tan(√𝑧)
√𝑧

𝑑𝑧

𝑑𝑧 = |𝑢 = √𝑧, 𝑑𝑢 = 2 𝑧| = 2 ∫ tan 𝑢 𝑑𝑢 = −2 ln|cos 𝑢| + 𝐶 = −𝟐 𝐥𝐧|𝐜𝐨𝐬(√𝒛)| + 𝐶.


𝑑𝑢

1

1

𝟏

15. ∫ 52𝑥 𝑑𝑥 = ∫ 𝑒 2𝑥 ln 5 𝑑𝑥 = |𝑢 = 2𝑥 ln 5 , 𝑑𝑢 = 2 ln 5 𝑑𝑥| = ∫ 𝑒 𝑢 2 ln 5 = 2 ln 5 𝑒 𝑢 + 𝐶 = 2 ln 5 𝑒 2𝑥 ln 5 + 𝐶 = 𝟐 𝐥𝐧 𝟓 𝟓𝟐𝒙 + 𝐶.

1

1

𝟏

16. ∫ 𝑒 3𝑥 sec(𝑒 3𝑥 ) 𝑑𝑥 = |𝑢 = 𝑒 3𝑥 , 𝑑𝑢 = 3𝑒 3𝑥 𝑑𝑥| = 3 ∫ sec 𝑢 𝑑𝑢 = 3 ln|tan 𝑢 + sec 𝑢| + 𝐶 = 𝟑 𝐥𝐧|𝐭𝐚𝐧(𝒆𝟑𝒙 ) + 𝐬𝐞𝐜(𝒆𝟑𝒙 )| + 𝐶.

𝑑𝑥

𝑑𝑥

17. ∫ 1+ 𝑥 = |𝑢 = 1 + √𝑥, 𝑑𝑢 = 2 𝑥 , 𝑑𝑥 = 2√𝑥𝑑𝑢 = 2(𝑢 − 1)𝑑𝑢| = ∫



2(𝑢−1)𝑑𝑢
𝑢

1

= 2 ∫ (1 − 𝑢) 𝑑𝑢 = 2(𝑢 − ln|𝑢|) + 𝐶 =

= 2(1 + √𝑥 − ln(1 + √𝑥)) + 𝐶 = 𝟐(√𝒙 − 𝐥𝐧(𝟏 + √𝒙)) +...


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