第 12 类 RD Sharma 解 - 第 19 章不定积分 - 练习 19.25 |设置 3

计算以下积分：

问题 41. ∫cos ^-1 ⁡((1 – x ² )/(1 + x ² ))dx

解决方案：

Given that, I = ∫cos^-1⁡((1 – x²)/(1 + x²))dx)

Let us considered x = tan⁡t

dx = sec²tdt

I = ∫cos^-1⁡((1 – tan²t)/(1 + tan²⁡t)) sec²tdt

= ∫cos^-1(cos⁡2t)sec²tdt

= ∫2tsec²⁡tdt

Using integration by parts,       

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

We get

I = 2[t∫sce²tdt – ∫(1∫sec²⁡tdt)dt]

= 2[t × tan²t – ∫tan⁡tdt]

= 2[t × tan²t – log⁡sec⁡t] + c

= 2[xtan^-1x – log⁡√(1 + x²)] + c

Hence, I = 2xtan^-1x – log⁡|1 + x²| + c

问题 42. ∫tan ^-1 ⁡(2x/(1 – x ² ))dx

解决方案：

Given that, I = ∫tan^-1⁡(2x/(1 – x²))dx

Let us considered x = tan⁡θ

dx = sec²θdθ

I = ∫tan^-1⁡((2tan⁡θ)/(1 – tan²θ)) sec²θdθ

= ∫tan^-1⁡(tan⁡2θ)sec²θdθ

= ∫2θsec²θdθ

Using integration by parts,       

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

We get

I = 2[θ∫sec²θdθ – ∫(1∫ sec²⁡θdθ)dθ]

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ – log⁡sec⁡θ] + c

= 2[xtan^-1⁡x – log⁡√(1 + x²)] + c

Hence, I = 2xtan^-1⁡x – log⁡|1 + x²| + c

问题 43. ∫(x + 1)log⁡xdx

解决方案：

Given that, I = ∫(x + 1)log⁡xdx

Using integration by parts,       

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

We get

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

= (x²/2 + x)log⁡x – ∫1/x (x²/2 + x)dx

= (x²/2 + x)log⁡x – 1/2 ∫xdx – ∫dx

= (x + x²/2)log⁡x – 1/2 × x²/2 – x + c

Hence, I = (x + x²/2)log⁡x – 1/2 × x²/2 – x + c

问题 44. ∫ x ² tan ^-1 xdx

解决方案：

Given that, I = ∫ x² tan^-1⁡xdx

Using integration by parts,       

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

We get

I = tan^-1⁡x∫x² dx – ∫(1/(1 + x²) ∫x² dx) dx

= tan^-1⁡x(x³/3) – 1/3∫x³/(1 + x²) dx

= 1/3 x³ tan^-1⁡x – 1/3 ∫(x – x/(1 + x²))dx

= 1/3 x³ tan^-1⁡x – 1/3 × x²/2 + 1/3 ∫x/(1 + x²) dx

Hence, I = 1/3 x³tan^-1x – 1/6 x² + 1/6 log⁡|1 + x²| + c

问题 45. ∫(e ^logx + sin⁡x) cos⁡xdx

解决方案：

Given that, I = ∫(e^logx + sin⁡x)cos⁡xdx

= ∫(x + sin⁡x)cos⁡xdx

= ∫xcos⁡xdx + ∫sin⁡xcos⁡xdx

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

= [xsin⁡x – ∫ sin⁡xdx] + 1/2 (-(cos⁡2x)/2) + c

I = xsin⁡x+cos⁡x – 1/4 cos⁡2x + c

= xsin⁡x + cos⁡x – 1/4 [1 – 2sin²⁡x] + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin²x + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin²x + c

Hence, I = xsin⁡x + cos⁡x + 1/2 sin²⁡x + d   [d = c-/4]

问题 46. ∫((xtan ^-1 ⁡x))/(1 + x ² ) ^3/2 dx

解决方案：

Given that, I = ∫((xtan^-1⁡x))/(1 + x²)^3/2dx

Let us considered tan^-1⁡x = t

1/(1 + x²) dx = dt

I = ∫(t tan⁡t)/√(1 + tan²⁡t) dt

= ∫(t × tan⁡t)/(sec⁡t) dt

= ∫t (sin⁡t)/(cos⁡t) cos⁡tdt

= ∫tsin⁡tdt

Using integration by parts,       

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

We get

I = [t]sin⁡tdt – ∫(1)sin⁡tdt)dt]

= [-tcos⁡t + ∫cos⁡tdt]

= [-tcos⁡t + sin⁡t] + c

= -(tan^-1⁡x)/√(1 + x²) + x/√(1 + x²) + c

Hence, I = -(tan^-1⁡x)/√(1 + x²) + x/√(1 + x²) + c

问题 47. ∫ tan ^-1 (√x)dx

解决方案：

Given that, I = ∫ tan^-1(√x)dx

Let us considered x = t²

dx = 2tdt

I = ∫2ttan^-1tdt

Using integration by parts,       

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

We get 

= 2[tan^-1)⁡t∫tdt – ∫(1/(1 + t²) ∫tdt)dt]

= 2[t²/2 tan^-1⁡t – ∫t²/2(1 + t²)dt]

= t² tan^-1⁡t – ∫(t² + 1 – 1)/(1 + t²)dt

= t² tan^-1⁡t – ∫(1 – 1/(1 + t²))dt

= t² tan^-1t – t + tan^-1⁡t + c

= (t² + 1) tan^-1⁡t – t + c

Hence, I = (x + 1)tan^-1⁡√x – √x + c

问题 48. ∫x ³ tan ^-1 xdx

解决方案：

Given that, I = ∫x³ tan^-1xdx

= tan^-1⁡x∫x³dx – (∫(dtan^-1⁡x)/dx (∫x³ dx)dx)

= tan^-1⁡x x⁴/4 – (∫1/(1 + x²) (x⁴/4)dx)

= tan^-1⁡x x⁴/4 – (∫1/(1 + x²) (x⁴/4)dx) 

= tan^-1⁡x x⁴/4 – (∫1/(1 + x²) (x⁴/4)dx)

∫ 1/(1 + x²) (x⁴/4)dx = 1/4 [∫1/(1 + x²) dx + (x² – 1)dx]

∫ 1/(1 + x²) (x⁴/4)dx = 1/4 [tan^-1⁡x + x³/3 – x]

Hence, I = x⁴/4 tan^-1⁡x – 1/4 [tan^-1⁡x + x³/3 – x] + c

问题 49. ∫xsin⁡xcos⁡2xdx

解决方案：

Given that, I = ∫xsin⁡xcos⁡2xdx

= 1/2 ∫x(2sin⁡xcos⁡2x)dx

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

= 1/2 ∫x(sin⁡3x – sin⁡x)dx

= 1/2[x](sin⁡3x – sin⁡x)dx – ∫ (1)(sin⁡3x – sin⁡x)dx)dx]

= 1/2 [x((-cos⁡3x)/3 + cos⁡x) – ∫(-(cos⁡3x)/3 + cos⁡x)dx]

Hence, I = 1/2 [-x (cos⁡3x)/3 + xcos⁡x + 1/9 sin⁡3x – sin⁡x] + c

问题 50. ∫(tan ^-1 x ² )xdx

解决方案：

Given that, I = ∫(tan^-1⁡x²)xdx

Let us considered x² = t

2xdx = dt

I = 1/2∫tan^-1tdt

= 1/2∫1tan^-1tdt

= 1/2 [tan^-1⁡t∫dt – (∫1/(1 + t²)∫dt)dt]

= 1/2 [t × tan^-1⁡t – ∫t/(1 + t²) dt]

= 1/2 t × tan^-1⁡t – 1/4∫2t/(1 + t²) dt

= 1/2 t × tan^-1⁡t – 1/4 log⁡|1 + t²| + c

Hence, I = 1/2 x² tan^-1⁡x² – 1/4 log⁡|1 + x⁴| + c

问题 51. ∫xdx/√(1 – x ² )

解决方案：

Given that, I = ∫xdx/√(1 – x²)

Let first function be sin^-1⁡x and second function be x/√(1 – x²).

Now, first we find the integral of the second function, 

∫xdx/√(1 – x²)

Now, put t = 1 – x²

Then dt = -2xdx

Therefore,

∫ xdx/√(1 – x²) = -1/2 ∫dt/√t = -√t = -√(1 – x²)

Hence,

∫(xsin^-1x)/√(1 – x²) dx

= (sin^-1⁡x)(-√(1 – x²) – ∫1/√(1 – x²) * (-√(1 – x²))dx

= -√(1 – x²) sin^-1⁡x + x + c

= x – √(1 – x²) sin^-1⁡x + c

问题 52. ∫sin ³ √x dx

解决方案：

Given that, I = ∫sin³√x dx

Let us considered √x = t

x = t²

dx = 2tdt

I = 2∫ tsin³⁡tdt

= 2∫t((3sin⁡t – sin⁡3t)/4)dt

= 1/2 ∫t(3sin⁡t – sin⁡3t)dt

Using integration by parts,       

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

We get

I = 1/2 [t(-3cos⁡t + 1/3 cos⁡3t) – ∫(-3cos⁡t + (cos⁡3t)/3)dt]

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 – {-3sin⁡t + (sin⁡3t)/9}] + c

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 + (27sin⁡t – 3sin⁡3t)/9] + c

= 1/18[-27tcos⁡t + 3tcos⁡3t + 27sin⁡t – 3sin⁡3t] + c

Hence, I = 1/18[3√x cos⁡3√x + 27sin⁡√x – 27√x cos⁡√x – 3sin⁡3√x] + c

问题 53. ∫ xsin ³ xdx

解决方案：

Given that, I = ∫ xsin³⁡xdx

= ∫x((3sin⁡x – sin⁡3x)/4)dx

= 1/4 ∫x(3sin⁡x – sin⁡3x)dx

Using integration by parts,       

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

We get

= 1/4 [x∫ (3sin⁡x – sin⁡3x)dx – ∫(1)(3sin⁡x – sin⁡3x)dx)dx]

= 1/4 [x(-3cos⁡x + (cos⁡3x)/3) – ∫(-3cos⁡x + (cos⁡3x)/3)dx]

= 1/4 [-3xcos⁡x + (xcos⁡3x)/3 + 3sin⁡x – (sin⁡3x)/9] + c

Hence, I = 1/36[3xcos⁡3x – 27xcos⁡x + 27sin⁡x – sin⁡3x] + c

问题 54. ∫cos ³ √x dx

解决方案：

Given that, I = ∫cos³√x dx

Let us considered x = t²

dx = 2tdt

= 2∫tcos³⁡tdt

= 2∫t((3cos⁡t + cos⁡3t)/4)dt

= 1/2 ∫t(3cos⁡t + cos⁡3t)dt

Using integration by parts,       

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

We get

I = 1/2 [t(3sin⁡t + 1/3 sin⁡3t) + ∫(1 × 3sin⁡t + (sin⁡3t)/3)dt]

= 1/2 [t((9sin⁡t + sin⁡3t)/3) + 3cos⁡t(cos⁡3t)/9] + c

= 1/18[27tsin⁡t + 3tsin⁡3t + 9cos⁡t + cos⁡3t] + c

Hence, I = 1/18[27√x sin⁡√x + 3√x sin⁡3√x + 9cos⁡√x + cos⁡3√x] + c

问题 55. ∫xcos ³ xdx

解决方案：

Given that, I = ∫xcos³⁡xdx

= ∫x((3cos⁡x + cos⁡3x)/4)dx

= 1/4 ∫x(3cos⁡x + cos⁡3x)dx

Using integration by parts,       

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

We get

I = 1/4 [x∫(3cos⁡x + cos⁡3x)dx – ∫(1)(3cos⁡x + cos⁡3x)dx)dx]

= 1/4 [x(3sin⁡x + (sin⁡3x)/3) – ∫ (3sin⁡x + (sin⁡3x)/3)dx]

= 1/4 [3xsin⁡x + (xsin⁡3x)/3 + 3cos⁡x + (cos⁡3x)/9] + c

Hence, I = (3xsin⁡x)/4 + (xsin⁡3x)/12 + (3cos⁡x)/4 + (cos⁡3x)/36 + c

问题 56. ∫tan ^-1 √((1 – x)/(1 + x))

解决方案：

Given that, I = ∫tan^-1√((1 – x)/(1 + x))

Let us considered x = cos⁡θ

dx = -sin⁡θdθ

I = ∫ tan^-1⁡(tan⁡θ/2)(-sin⁡θ)dθ

=-1/2 ∫θsin⁡θdθ

Let θ = u and sin⁡θdθ = v 

So that sin⁡θ = ∫vdθ

Using integration by parts,       

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

We get

I = -1/2 (-θcos⁡θ – ∫-cos⁡θdθ)

= -1/2(-θcos⁡θ + sin⁡θ)+c

= -1/2 (-θcos⁡θ + √(1 – cos²⁡θ)) + c

= -1/2 (-xcos^-1⁡x + √(1 – x²)) + c

问题 57. ∫sin ^-1 √(x/(a + x)) dx

解决方案：

Given that, I = ∫sin^-1⁡√(x/(a + x)) dx

Let us considered x = atan²θ

dx = 2atan⁡θsec²⁡θdθ

I = ∫(sin^-1⁡√((atan²⁡θ)/(a + atan²⁡θ))(2atan⁡θsec²θ)dθ

= ∫ (sin^-1√((tan²θ)/(sec²θ)))(2atan⁡θsec²θ)dθ

= ∫ sin^-1(sin⁡θ)(2atan⁡θsec²θ)dθ

= ∫ 2θatan⁡θsec²θdθ

= 2a∣θ(tan⁡θsec²⁡θ)dθ)

= ∫2θatan⁡θsec²θdθ

= 2a∫θ(tan⁡θsec²⁡θ)dθ

= 2a[θ]tan⁡θsec²θdθ – ∫(∫tan⁡θsec²⁡θdθ)dθ]

= 2a[θ (tan²⁡θ)/2 – ∫(tan²θ)/2 dθ]

= aθtan²θ – 2a/2∫(sec²θ – 1)dθ

= aθtan²θ – atan⁡θ + aθ + c

= a(tan^-1⁡√(x/a)) x/a – a√(x/a) + atan^-1⁡√(x/a) + c

Hence, I = xtan^-1⁡√(x/a) – √ax + atan^-1⁡√(x/a) + c

问题 58. ∫(x ³ sin ^-1 ⁡x²)/√(1 – x ⁴ ) dx

解决方案：

Given that, I = ∫(x³ sin^-1x²)/√(1 – x⁴) dx

Let us considered sin^-1⁡x² = t

(1/√(1 – x⁴)(2x)dx = dt

I = ∫(x² sin^-1⁡x²)/√(1 – x⁴) xdx

= ∫(sin⁡t)t dt/2

= 1/2∫tsin⁡tdt

= 1/2 [t∫sin⁡tdt – ∫(1∫sin⁡tdt)dt]

= 1/2 [t(-cost)dt – ∫(1∫(-cost))dt]

= 1/2[-tcost + sint] + c

Hence, I = 1/2 [x² – √(1 – x⁴) sin^(-1)⁡x²] + c

问题 59. ∫(x ² sin ^-1 ⁡x)/(1 – x ² ) ^3/2 dx

解决方案：

Given that, I = ∫(x² sin^-1x)/(1 – x²)^3/2dx

Let us considered sin^-1⁡x = t

(1/√(1 – x²) dx = dt

I = ∫(sin²t × t)/((1 – sin²t)) dt

= ∫(tsin²t)/(cos²t) dt

= ∫t × tan²tdt

= ∫t(sec²⁡t – 1)dt

= ∫tsec²⁡tdt – t²/2 + c

= t∫sec²tdt – ∫(1∫sec²tdt)dt – t²/2 + c

= t × tan⁡t – ∫tan⁡tdt – t²/2 + c

= t × tan⁡t – log⁡sec⁡t – t²/2 + c

Hence, I = x/√(1 – x²) sin^-1x + log⁡|1 – x²| – 1/2 (sin^-1x)² + c

问题 60. ∫cos ^-1 (1 – x ² / 1 + x ² ) dx

解决方案：

Given that, I = ∫cos^-1(1 – x²/ 1 + x²) dx

Let us considered, x = tant

dx = sec²tdt

I = ∫cos^-1(1 – tan²t/ 1 + tan²t) sec²tdt

= ∫ 2t sec²tdt

Using integration by parts,       

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

We get

I = 2[t∫sec²tdt – ∫(1 ∫sec²tdt)dt]

= 2[t tan²t – ∫tant dt]

= 2[t tan²t – log sect] + c

= 2[x tan²x – log √1 + x²] + c

Hence, I = 2[xtan²x – log √1 + x²] + c