第 12 类 RD Sharma 解 - 第 19 章不定积分 - 练习 19.25 |设置 2

计算以下积分：

问题 21. ∫(log⁡x) ² x dx

解决方案：

Given that, I = ∫(log⁡x)² x dx

Using integration by parts,       

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

We get

I = (log⁡x)²∫xdx – ∫(2(log⁡x)(1/x) ∫xdx)dx

= x²/2(log⁡x)² – 2∫(log⁡x)(1/x)(x²/2)dx

= x²/2(log⁡x)² – ∫x(log⁡x)dx

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

= x²/2(log⁡x)² – [x²2/2 log⁡x – ∫(1/x × x²/2)dx]

= x²/2(log⁡x)² – x²/2 log⁡x + 1/2 ∫xdx

= x²/2(log⁡x)² – x²/2 log⁡x + 1/4 x² + c

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

问题 22. ∫e ^√x dx

解决方案：

Given that, I = ∫ e^√x dx

 Let us assume, √x = t

x = t²

dx = 2tdt

I = 2∫ e^t tdt

Using integration by parts,       

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

We get

I = 2[t∫e^tdt – ∫(1∫e^tdt)dt]

= 2[te^t – ∫e^t dt]

= 2[te^t – e^t] + c

= 2e^t (t – 1) + c

Hence, I = 2e^√x(√x – 1) + c

问题 23. ∫(log⁡(x + 2))/((x + 2) ² ) dx

解决方案：

Given that, I = ∫(log⁡(x + 2))/((x + 2)²) dx

 Let us assume (1/(x + 2) = t

-1/((x + 2)²) dx = dt

I = -∫log⁡(1/t)dt

= -∫log⁡t^-1 dt

= -∫1 × log⁡tdt

Using integration by parts,       

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

We get

I = log⁡t∫dt – ∫(1/t ∫dt)dt

= tlog⁡t – ∫(1/t × t)dt

= tlog⁡t – ∫dt

= tlog⁡t – t + c

= 1/(x + 2) (log⁡(x + 2)^-1 – 1) + c

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

问题 24. ∫(x + sin⁡x)/(1 + cos⁡x) dx

解决方案：

Given that, I = ∫(x + sin⁡x)/(1 + cos⁡x) dx

= ∫x/(2cos²x/2) dx + ∫(2sin⁡x/2 cos⁡x/2)/(2cos²⁡x/2) dx

= 1/2 ∫xsec²⁡x/2 dx + ∫tan⁡x/2 dx

Using integration by parts,       

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

We get

= 1/2 [x∫sec²x/2 dx – ∫(1∫ sec²x/2 dx)dx] + ∫tan⁡x/2 dx

= 1/2 [2xtan⁡x/2 – 2∫tan⁡x/2 dx] + ∫tan⁡x/2 dx + c

= xtan⁡x/2 – ∫tan⁡x/2 dx + ∫tan⁡x/2 dx+c

Hence, I = xtan⁡x/2 + c

问题 25. ∫log ₁₀ xdx

解决方案：

Given that, I = ∫log₁₀⁡xdx

= ∫(log⁡x)/(log⁡10) dx

= 1/(log⁡10) ∫1 × log⁡xdx

Using integration by parts,       

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

We get

= 1/(log⁡10) [log⁡x∫dx – ∫(1/x ∫dx)dx]

= 1/(log⁡10) [xlog⁡x – ∫(x/x)dx]

= 1/(log⁡10)[xlog⁡x – x]

Hence, I = (x/(log⁡10)) × (log⁡x – 1) + c

问题 26. ∫cos⁡√x dx

解决方案：

Given that, I = ∫cos⁡√x dx

Let us assume, √x = t

x = t²

dx = 2tdt

= ∫2tcos⁡tdt

I = 2∫tcos⁡tdt

Using integration by parts,       

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

We get

I = 2[t]cos⁡tdt – ∫(1 ∫cos⁡tdt)dt]

= 2[tsin⁡t – ∫sin⁡tdt]

= 2[tsin⁡t + cos⁡t] + c

Hence, I = 2[√x sin⁡√x + cos√x] + c

问题 27. ∫(xcos ^-1 x)/√(1 – x ² ) dx

解决方案：

Given that, I = ∫(xcos^-1x)/√(1 – x²) dx

Let us assume, t = cos^-1⁡x

dt = (-1)/√(1 – x²) dx

Also, cost = x

I = -∫tcos⁡tdt

Now, using integration by parts,       

So, let

u = t;

du = dt

∫cos⁡tdt = ∫dv

sin⁡t = v

Therefore,

 I = -[tsint – ∫sin⁡tdt]

= -[tsint + cos⁡t] + c

On substituting the value t = cos^-1x we get,

 I = -[cos^-1⁡xsin⁡t + x] + c

Hence, I = -[cos^-1⁡x√(1 – x²) + x] + c

问题 28. ∫cosec ³ xdx

解决方案：

Given that, I =∫cosec³xdx

 =∫cosec⁡x × cosec²xdx

Using integration by parts,       

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

We get

= cosec⁡x × ∫cosec²xdx + ∫(cosec⁡xcot⁡x]cosec²xdx)dx

= cosecx × (-cot⁡x) + ∫cosec⁡xcot⁡x(-cot⁡x)dx

= -cosec⁡xcot⁡x – ∫cosecx⁡cot²⁡xdx

= -cosec⁡xcot⁡x – ∫cosec⁡x(cosec²x – 1)dx

= -cosec⁡xcot⁡x – ∫cosec³xdx + ∫cosecxd⁡x

I = -cosec⁡xcot⁡x – I + log⁡|tan⁡x/2| + c₁

2l = -cosec⁡xcot⁡x + log⁡|tan⁡x/2| + c₁

Hence, I = -1/2cosecx⁡cot⁡x + 1/2 log⁡|tan⁡x/2| + c

问题 29. ∫sec ^-1 √x dx

解决方案：

Given that, I = ∫sec^-1⁡√x dx

 Let us assume, √x = t

x = t²

dx = 2tdt

I = ∫2tsec^-1⁡tdt

Using integration by parts,       

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

We get

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

= 2[t²/2 sec^-1 – ∫(t/(2t√(t² – 1)))dt]

= t² sec^-1⁡t – ∫t/√(t² – 1) dt

= t² sec^-1⁡t – 1/2∫2t/√(t² – 1) dt

= t² sec^-1⁡t – 1/2 × 2√(t² – 1) + c

Hence, I = xsec^-1⁡√x – √(x – 1) + c

第 30 题 ∫sin ^-1 √x dx

解决方案：

Given that, I = ∫sin^-1⁡√x dx 

Let us assume, x = t

dx = 2tdt

∫sin^-1√x dx = ∫sin^-1⁡√(t²) 2tdt

= ∫sin^-1⁡t2tdt

= sin⁡^-1t∫2tdt – (∫(dsin^-1⁡t)/dt (∫2tdt)dt

= sin^-1⁡t(t²) – ∫1/√(1 – t²) (t²)dt

Now, lets solve ∫1/√(1 – t²) (t²)dt

∫1/√(1 – t²) (t²)dt = ∫(t² – 1 + 1)/√(1 – t²) dt

= ∫(t² – 1)/√(1 – t²) dt + ∫1/√(1 – t²) dt

As we know that, value of ∫1/√(1 – t²) dt = sin^-1⁡t

So, the remaining integral to evaluate is 

∫(t² – 1)/√(1 – t²) dt= ∫-√(1 – t²) dt

Now, substitute, t = sin⁡u, dt = cos⁡udu, we gte

∫-√(1 – t²) dt = ∫-cos²udu = -∫[(1 + cos⁡2u)/2]du

= -u/2-(sin⁡2u)/4

Now substitute back u = sin^-1x and t = √x, we get

= -(sin^-1√x)/2 – (sin⁡(2sin^-1⁡√x))/4

∫sin^-1√x dx = xsin^-1⁡√x-(sin^-1⁡√x)/2 – (sin⁡(2sin^-1√x))/4

sin⁡(2sin^-1⁡√x) = 2√x √(1 – x)

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

问题 31. ∫xtan ² ⁡xdx

解决方案：

Given that, I =∫xtan²⁡xdx

 = ∫x(sec²x – 1)dx

Using integration by parts,       

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

We get

 = ∫xsec8xdx – ∫xdx

 = [x∫sec²xdx – ∫(1∫sec²xdx)dx] – x²/2

 = xtan⁡x – ∫tan⁡xdx – x²/2

Hence, I = xtan⁡x – log⁡|sec⁡x| – x²/2 + c

问题 32. ∫ x((sec⁡2x – 1)/(sec⁡2x + 1))dx

解决方案：

Given that, I = ∫ x((sec⁡2x – 1)/(sec⁡2x + 1))dx

= ∫x((1 – cos⁡2x)/(1 + cos⁡2x))dx

= ∫x((sec²⁡x)/(cos²⁡x))dx

= ∫xtan²xdx

= ∫x(sec²⁡x – 1)dx

Using integration by parts,       

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

We get

= ∫xsec²xdx – ∫dx

= [x∫sec²⁡xdx – ∫(1∫  sec²xdx)dx] – x²/2

= xtan⁡x – ∫tan⁡xdx – x²/2

= xtan⁡x – log|secx| – x²/2 + c

Hence, I = xtan⁡x – log|secx| – x²/2 + c

问题 33. ∫(x + 1)e ^x log⁡(xe ^x )dx

解决方案：

Given that, I = ∫(x + 1)e^xlog⁡(xe^x)dx

Let us assume, xe^x = t

(1 × e^x + xe^x)dx = dt

(x + 1)e^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 × t)dt

= tlog⁡t – ∫dt

= tlog⁡t – t + c

= t(log⁡t – 1) + c

Hence, I = xe^x (log⁡xe^x – 1) + c

问题 34. ∫sin ^-1 (3x – 4x ³ )dx

解决方案：

Given that, I = ∫sin^-1(3x – 4x³)dx

Let us assume, x = sin⁡θ

dx = cos⁡θdθ

= ∫sin^-1⁡(3sin⁡θ – 4sin³⁡θ)cos⁡θdθ

= ∫sin^-1(sin⁡3θ)cos⁡θdθ

= ∫3θcos⁡θdθ

Using integration by parts,       

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

We get

= 3[θ]cos⁡θdθ – ∫(1∫cos⁡θdθ)dθ]

= 3[θsin⁡θ – ∫sin⁡θdθ]

= 3[θsin⁡θ + cos⁡θ] + c

Hence, I = 3[xsin^-1⁡x + √(1 – x²)] + c)

问题 35. ∫sin ^-1 (2x/(1 + x ² ))dx。

解决方案：

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

Let us assume, x = tan⁡θ

dx = sec²⁡θdθ

sin^-1⁡(2x/(1 + x²)) = sin^-1⁡((2tan⁡θ)/(1 + tan²⁡θ))

= sin^-1⁡(sin⁡2θ) = 2θ

∫sin^-1⁡(2x/(1 + x²))dx = ∫2θsec²θdθ = 2∫θsec²⁡θdθ

Using integration by parts,       

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

We get

2[θ∫sec²⁡θdθ – ∫{(d/dθ θ) ∫sec²θdθ}dθ

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ + log⁡|cos⁡θ|] + c

= 2[xtan^-1⁡x + log⁡|1/√(1 + x²)|] + c

= 2xtan^-1⁡x + 2log⁡(1 + x²)^1/2 + c

= 2xtan^-1⁡x + 2[-1/2 log⁡(1 + x²)] + c

= 2xtan^-1⁡x – log⁡(1 + x²) + c

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

问题 36. ∫ tan ^-1 ((3x – x ³ )/(1 – 3x ² ))dx

解决方案：

Given that, I = ∫tan^-1⁡((3x – x³)/(1 – 3x²))dx

 Let us assume, x = tan⁡θ

dx = sec²⁡θdθ

I = ∫tan^-1⁡((3tan⁡θ – tan³θ)/(1 – 3tan²⁡x)) sec²⁡θdθ

=∫tan^-1⁡(tan⁡3θ)sec²θdθ

= ∫3θsec²θdθ

Using integration by parts,       

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

We get

= 3[θ∫ sec²⁡θdθ – ∫(1∫sec²θdθ)dθ]

= 3[θtan⁡θ – ∫tan⁡θdθ]

= 3[θtan⁡θ + log⁡sec⁡θ] + c

= 3[xtan^-1x – log⁡√(1 + x²)] + c

Hence, I = 3[xtan^-1x – log⁡√(1 + x²)] + c

问题 37. ∫x ² sin ^-1 xdx

解决方案：

Given that, I = ∫x²sin^-1⁡xdx

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

= x³/3 sin^-1⁡x – ∫x³/(3√(1 – x²)) dx

I = x³/3 sin^-1⁡x – 1/3 I₁ + c₁ …..(1)

Let I₁ = ∫x³/√(1 – x²) dx

Let 1 – x² = t²

-2xdx = 2tdt

-xdx = tdt

I₁ = -∫(1 – t²)tdt/t

= ∫(t² – 1)dt

= t³/3 – t + c₂

= (1 – x²)^3/2/3 – (1 – x²)^1/2 + c₂

Now, put the value of I₁ in eq(1), we get

Hence, I = x³/3 sin^-1⁡x – 1/9 (1 – x²)^3/2 + 1/3 (1 – x²)^1/2 + c

问题 38. ∫(sin ^-1 x)/x ² dx

解决方案：

Given that, I =∫(sin^-1⁡x)/x²dx

 = ∫(1/x²)(sin^-1⁡x)dx

I = [sin^-1⁡x∫1/x²dx – ∫(1/√(1 – x²) ∫1/x²dx)dx]

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

I = -1/x sin^-1x + ∫1/(x√(1 – x²)) dx

I = -1/x sin^-1⁡x + I₁  …….(1)

Where,

I₁ = ∫1/(x√(1 – x²)) dx

Let 1 – x² = t²

-2xdx = 2tdt

I₁ = ∫x/(x²√(1 – x²)) dx

= -∫tdt/((1 – t²) √t)

= -∫dt/((1 – t²))

= ∫1/(t² – 1) dt

= 1/2 log⁡|(t – 1)/(t + 1)| 

= 1/2 log⁡|(t – 1)/(t + 1)|

= 1/2 log⁡|(√(1 – x²) – 1)/(√(1 – x²) + 1)| + c₁

Now, put the value of I₁ in eq(1), we get

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

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

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

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

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

问题 39. ∫(x ² tan ^-1 x)/(1 + x ² ) dx

解决方案：

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

Let us assume, tan^-1⁡x = t     [x = tan⁡t]

1/(1 + x²) dx = dt

I = ∫t × tan²tdt

= ∫t(sec²⁡t – 1)dt

= ∫(tsec²t – t)dt

Using integration by parts,       

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

We get

= ∫tsec²tdt – ∫tdt

= [t∫sec²tdt – ∫(1)sec²⁡tdt)dt] – t²/2

= [t × tan⁡t – ∫tan⁡tdt] – t²/2

= t tan⁡t – log⁡sec⁡t – t²/2 + c

= xtan^-1⁡x – log⁡√(1 + x²) – (tan²x)/2 + c

Hence, I = xtan^-1⁡x – 1/2 log⁡|1 + x²| – (tan²⁡x)/2 + c

问题 40. ∫cos ^-1 (4x ³ – 3x)dx

解决方案：

Given that, I = ∫cos^-1⁡(4x³ – 3x)dx

 Let us assume, x = cos⁡θ

dx = -sin⁡θdθ

I = -∫cos^-1(4cos⁡³θ – 3cos⁡θ)sin⁡θdθ

= – ∫cos^-1(cos⁡3θ)sin⁡θdθ

= -∫3θsin⁡θdθ

Using integration by parts,       

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

We get

= -3[θ]sin⁡θdθ – ∫(1∫sin⁡θdθ)dθ]

= -3[-θcos⁡θ + ∫cos⁡θdθ]

= 3θcos⁡θ – 3sin⁡θ + c

Hence, I = 3xcos^-1x – 3√(1 – x²) + c