第 12 类 RD Sharma 解 - 第 19 章不定积分 - 练习 19.25 |设置 1

计算以下积分：

问题 1. ∫x cos⁡xdx

解决方案：

Given that, I = ∫x cos⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫cos⁡xdx – ∫(1 × ∫cos⁡xdx)dx + c

= xsin⁡x – ∫sin⁡xdx + c

Hence, I = x sin⁡x + cos⁡x + c

问题 2. ∫log⁡(x + 1)dx

解决方案：

Given that, I = ∫log⁡(x + 1)dx

= ∫1 × log⁡(x + 1)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡(x + 1)∫1dx – ∫(1/(x + 1) × ∫ 1dx)dx + c

= xlog⁡(x + 1) – ∫(x/(x + 1))dx + c

= x log⁡(x + 1) – ∫(1 – 1/(x + 1))dx + c

Hence, I = x log⁡(x + 1) – x + log⁡(x + 1) + c

问题 3. ∫x ³ log⁡xdx

解决方案：

Given that, I = ∫ x³ log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x ∫x³ dx – ∫(1/x × ∫x³ dx)dx + c

= x⁴/4 log⁡x – ∫x⁴/4x dx+c

= x⁴/4 log⁡x – 1/4∫x³ dx + c

= x⁴/4 log⁡x – 1/4 ∫x⁴/4 dx + c

I = x⁴/4 log⁡x – 1/16 x⁴ + c

问题 4. ∫xe ^x dx

解决方案：

Given that I = ∫xe^x dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = xe^x – ∫1.e^x dx

= xe^x – e^x + c

Hence, I = = xe^x – e^x + c

问题 5. ∫xe ^2x dx

解决方案：

Given that, I = ∫xe^2x dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫e^2x dx – ∫(1 × ∫ e^2x dx) dx + c

= x∫e^2x dx – ∫(1 × ∫e^2x dx)dx + c

= (xe^2x)/2 – ∫(e^2x/2)dx + c

= (xe^2x)/2 – e^2x/4 + c

Hence, I = (x/2 – 1/4) e^2x + c

问题 6. ∫x ² e ^-x dx

解决方案：

Given that I = ∫x² e^-x dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x² ∫e^-x dx – ∫(2x∫e^-x dx)dx

= -x² e^-x – ∫(2x)(-e^-x)dx

= -x² e^-x + 2∫xe^-x dx

= -x² e^-x + 2[x∫e^-x dx – ∫(1 × ∫ e^-x dx) dx]

= -x² e^-x + 2[x(-e^-x) – ∫(-e^-x)dx]

= -x² e^-x – 2xe^-x + 2∫e^-x dx

Hence, I = -x² e^-x – 2xe^-x – 2e^-x + c

问题 7. ∫ x ² cos⁡xdx

解决方案：

Given that, I = ∫ x²cos⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

 I = x² ∫ cos⁡xdx – ∫(2x)cos⁡xdx)dx

= x² sin⁡x – 2∫(x)(sin⁡x)dx

= x² sin⁡x – 2[x∫sin⁡xdx – ∫(1 × ∫sin⁡xdx)dx]

= x² sin⁡x – 2[x(-cos⁡x) – ∫(-cos⁡x)dx]

= x² sin⁡x + 2xcos⁡x – 2∫(cos⁡x)dx

Hence, I = x²sin⁡x + 2xcos⁡x – 2sin⁡x + c

问题 8. ∫x ² cos⁡2xdx

解决方案：

Given that, I = ∫x²cos⁡2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x² ∫cos⁡2xdx – ∫(2x∫ cos⁡2xdx)dx

= x² (sin⁡2x)/2 – 2∫x((sin⁡2x)/2)dx

= 1/2 x² sin⁡2x – ∫xsin⁡2xdx

= 1/2 x² sin⁡2x – [x∫sin⁡2xdx – ∫ (1∫ sin⁡2xdx)dx]

= 1/2 x² sin⁡2x – [x((-cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx]

= 1/2 x²sin⁡2x + x/2 cos⁡2x – 1/2 ∫(cos⁡2x)dx

Hence, I = 1/2 x² sin⁡2x + x/2 cos⁡2x – 1/4 sin⁡2x + c

问题 9. ∫xsin⁡2xdx

解决方案：

Given that, I =∫xsin⁡2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

 I = x∫sin⁡2xdx – ∫(1)sin⁡2xdx)dx

= x(-(cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx

= -x/2 cos⁡2x + 1/2 ∫cos⁡2xdx

= -x/2 cos⁡2x + 1/2(sin⁡2x)/2 + c

Hence, I = -x/2 cos⁡2x + 1/4 sin⁡2x + c

问题 10. ∫(log⁡(log⁡x))/x dx

解决方案：

 Given that, I = ∫(log⁡(log⁡x))/x dx 

= ∫(1/x)(log⁡(log⁡x))dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡log⁡x]1/x dx – ∫(1/(xlog⁡x)∫1/x dx)dx

= log⁡x × log⁡(log⁡x) – ∫(1/(xlog⁡x) log⁡x)dx

= log⁡x × log⁡(log⁡x) – ∫1/x dx

= log⁡x × log⁡(log⁡x) – log⁡x + c

Hence, I = log⁡x(log⁡log⁡x – 1) + c

问题 11. ∫x ² cos⁡xdx

解决方案：

Given that I = ∫x² cos⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x²∫ cos⁡xdx – ∫(2x]cos⁡xdx)dx

= x²sin⁡x – 2∫xsin⁡xdx

= x² sin⁡x – 2[x∫sin⁡xdx – ∫(1]sin⁡xdx)dx]

= x² sin⁡x – 2[x(-cos⁡x) – ∫(-cos⁡x)dx]

= x² sin⁡x + 2xcos⁡x – 2∫(cos⁡x)dx

Hence, I = x² sin⁡x + 2xcos⁡x – 2sin⁡x + c

问题 12. ∫xcosec ² ⁡xdx

解决方案 ：

Given that, I = ∫xcosec²⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫cosec²xdx – ∫(∫ cosec²xdx)dx

= -xcot⁡x + ∫cot⁡xdx

= -x cot⁡x + log ⁡|sin⁡x| + c

Hence, I = -x cot⁡x + log ⁡|sin⁡x| + c

问题 13. ∫xcos ² xdx

解决方案：

Given that, I = ∫xcos²xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = x∫cos²⁡xdx – ∫(1∫ cos²xdx)dx

= x∫((cos⁡2x + 1)/2)dx – ∫(∫((1 + cos⁡2x)/2)dx)dx

= x/2 [(sin⁡2x)/2 + x] – 1/2∫(x + (sin⁡2x)/2)dx

= x/4 sin⁡2x + x²/2 – 1/2 × x²/2 – 1/4 (-(cos⁡2x)/2) + c

Hence, I = x/4 sin⁡2x + x²/4 + 1/8 cos⁡2x + c

问题 14. ∫x ⁿ log⁡x dx

解决方案：

Given that, I = ∫xⁿ log⁡xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x∫xⁿ dx – ∫(1/x ∫xⁿdx)dx

= xⁿ⁺¹/(n + 1) log⁡x – ∫(1/x × xⁿ⁺¹/(n + 1))dx

= xⁿ⁺¹/(n + 1) log⁡x – ∫(xⁿ/(n + 1))dx

Hence, I = xⁿ⁺¹/(n + 1) log⁡x – 1/(n + 1)² × (xⁿ⁺¹) + c

问题 15. ∫(log⁡x)/x ⁿ dx

解决方案：

Given that, I = ∫(log⁡x)/xⁿ dx = ∫(log⁡x)(1/xⁿ)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

I = log⁡x∫(1/xⁿ)dx – ∫((d(log⁡x))/dx)(∫(1/xⁿ)dx)dx

= log⁡x(x^1-n/(1 – n)) – ∫1/x (x^1-n/(1 – n))dx

= log⁡x(x^1-n/(1 – n)) – ∫(xⁿ/(1 – n))dx

= log⁡x(x^1-n/(1 – n)) – (1/(1 – n))(x^1-n/(1 – n))

Hence, I = log⁡x(x^1-n/(1 – n)) – (x^1-n/([1 – n]²)) + c

问题 16. ∫x ² sin ² ⁡xdx

解决方案：

Given that, I = ∫x² sin²⁡xdx

= ∫x² ((1 – cos⁡2x)/2)dx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= ∫x²/2 dx – ∫((x² cos⁡2x)/2)dx

= x³/6 – 1/2 [∫x² cos⁡2xdx]

= x³/6 – 1/2 [x² ∫cos⁡2xdx – ∫ (2x∫cos⁡2xdx)dx]

= x³/6 – 1/2 (x²(sin⁡2x)/2) + 1/2 × 2∫(x (sin⁡2x)/2)dx

= x³/6 – 1/4 x²sin⁡2x + 1/2 [x ∫sin⁡2xdx – ∫(1∫sin⁡2xdx)dx]

= x³/6 – 1/4 x² sin⁡2x + 1/2 [x(-(cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx] 

= x³/6 – 1/4 x² sin⁡2x + 1/2 x(-(cos⁡2x)/2) + 1/4 × (sin2x/2) + c

= x³/6 – 1/4 x² sin⁡2x – 1/4 x(cos⁡2x) + 1/8 × (sin2x) + c

Hence, I = x³/6 – 1/4 x² sin⁡2x – 1/4 x(cos⁡2x) + 1/8 × (sin2x) + c

问题 17。 $∫2x^3 e^{x^2} xdx$

解决方案：

Given that, l = $∫2x^3 e^{x^2} xdx$

 Let us assume, x² = t

2xdx = dt

I = ∫t × e^t dt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= t∫e^t dt – ∫(1 × ∫e^tdt)dt

= te^t – ∫e^t dt

= te^t – e^t + c

= e^t-1 + c

Hence, I = $e^{x^2}$ (x² – 1) + c

问题 18. ∫x ³ cos⁡x ² dx

解决方案：

Given that, I = ∫x³ cos⁡x² dx

Let us assume x² = t

2xdx = dt

I = 1/2 ∫tcos⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/2[t∫cos⁡tdt – ∫(1 × ∫cos⁡tdt)dt]

= 1/2 [t × sin⁡t – ∫sin⁡tdt]

= 1/2[tsin⁡t + cos⁡t] + c

Hence, I = 1/2 [x² sin⁡x² + cos⁡x²] + c

问题 19. ∫xsin⁡xcos⁡xdx

解决方案：

Given that, I = ∫xsin⁡xcos⁡xdx

 = ∫x/2(2sin⁡xcos⁡x)dx

 = 1/2 ∫xsin⁡2xdx

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= 1/2 [x∫sin⁡2xdx – ∫(1 × ∫sin⁡2xdx)dx]

= 1/2 [x((-cos⁡2x)/2) – ∫((-cos⁡2x)/2)dx]

= -1/4 xcos⁡2x + 1/4 ∫cos⁡2xdx

Hence, I = -1/4 xcos⁡2x + 1/8 sin⁡2x + c

问题 20. ∫sin⁡x(log⁡cos⁡x)dx

解决方案：

Given that, I = ∫sin⁡x(log⁡cos⁡x)dx

 Let us considered, cos⁡x = t

 -sin⁡xdx = dt

I = -∫ log⁡tdt

 = -∫1 × log⁡tdt

Using integration by parts,       

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c 

We get

= -[log⁡t∫dt – ∫(1/t × ∫dt)dt]

= -[tlog⁡t – ∫1/t × tdt]

= -[tlog⁡t-∫  dt]

= -[tlog⁡t – t + c₁ ]

= t(1 – logt) + c

Hence, I = cosx(1 – logcosx) + c