第12类RD Sharma解–第19章不定积分–练习19.9 |套装3

📌 相关文章

📜 第12类RD Sharma解–第19章不定积分–练习19.9 |套装3

📅 最后修改于: 2021-06-24 19:30:31 🧑 作者: Mango

问题49。 $∫5^{x + tan^{-1}x}(\frac{x^2 + 2}{x^2 + 1})dx$

解决方案：

Given that I = $∫5^{x + tan^{-1}x}(\frac{x^2 + 2}{x^2 + 1})dx$ ……..(i)

Let us considered x + tan^-1⁡x = t then,

On differentiating both side we get,

d(x + tan^-1⁡x) = dt

(1 + 1/(1 + x²))dx = dt

((1 + x²+ 1)/(1 + x²))dx = dt

((x²+ 2))/((x²+ 1)) dx = dt

Now on putting x + tan^-1⁡x = t and ((x²+ 2)/(x²+ 1))dx = dt in equation (i), we get

I = ∫5^tdt

= 5^t/(log⁡5) + c

= $5^{x + tan^{-1}x}$ /(log⁡5) + c

Hence, I = $5^{x + tan^{-1}x}$ /(log⁡5) + c

为什么编程需要懂一点英语

问题50。 $∫\frac{e^{msin^{-1}x}}{\sqrt{1-x^2}} dx$

解决方案：

Given that I = $∫\frac{e^{msin^{-1}x}}{\sqrt{1-x^2}} dx$ ……(i)

Let us considered msin^-1⁡x = t then,

On differentiating both side we get,

d(msin^-1x) = dt

m 1/√(1 – x²) dx = dt

dx/√(1 – x²) = dt/m

Now on putting msin^-1⁡x = t and dx/√(1 – x²) = dt/m in equation (i), we get

I = ∫e^tdt/m

= 1/m e^t+c

= $\frac{1}{m}e^{msin^{-1}x}+c$

Hence, I = $\frac{1}{m}e^{msin^{-1}x}+c$

为什么编程需要懂一点英语

问题51.∫(cos⁡√x)/√xdx

解决方案：

Given that I = ∫(cos⁡√x)/√x dx

Let us considered √x = t then,

On differentiating both side we get,

1/(2√x) dx = dt

Now,

= ∫ (cos⁡√x)/√x dx

= 2∫ cos⁡tdt2

= 2sin⁡t + c

Hence, I = 2sin⁡√x + c

为什么编程需要懂一点英语

问题52.∫sin(tan ^-1 x)/(1 + x ² )dx

解决方案：

Given that I = ∫sin(tan^-1x)/(1 + x²) dx ……(i)

Let us considered tan^-1= t then,

On differentiating both side we get,

d(tan^-1⁡x) = dt

1/(1 + x²) dx = dt

Now on putting tan^-1x = t and dx/(1 + x²) = dt in equation (i), we get

I = ∫ sin⁡tdt

= -cos⁡t + c

= -cos⁡(tan^-1x) + c

Hence, I = -cos⁡(tan^-1⁡x) + c

为什么编程需要懂一点英语

问题53.∫(sin⁡(log⁡x))/ x dx

解决方案：

Given that I = ∫(sin⁡(log⁡x))/x dx ……..(i)

Let us considered log⁡x = t then,

On differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

Now on putting log⁡x = t and 1/x dx = dt in equation (i), we get

I = ∫sin⁡tdt

= -cos⁡t + c

= -cos⁡(log⁡x) + c

Hence, I = -cos⁡(log⁡x) + c

为什么编程需要懂一点英语

问题54。 $∫\frac{e^{mtan^{-1}x}}{(1+x^2)}dx$

解决方案：

Given that I = $∫\frac{e^{mtan^{-1}x}}{(1+x^2)}dx$

Let us considered tan^-1⁡x = t, then

On differentiating the above function we have,

1/(1 + x²) dx = dt

$∫\frac{e^{mtan^{-1}x}}{(1+x^2 )} dx$ = ∫e^mt× dt

= $∫\frac{e^{mtan^{-1}x}}{(1+x^2 )} dx$ = e^mt/m

On Substituting the value of t, we

I = $\frac{e^{mtan^{-1}x}}{(m)}$ + c

为什么编程需要懂一点英语

问题55.∫x/(√(x ² + a ² )+√(x ² – a ² ))dx

解决方案：

Given that I = ∫x/(√(x²+ a²) + √(x²– a²)) dx

= ∫ x/(√(x² + a²) + √(x² – a²)) × (√(x² + a²) – √(x² – a² ))/(√(x²+ a²) – √(x² – a²)) dx

= ∫ x(√(x²+ a²) – √(x²– a²))/(x²+ a²– x²+ a²) dx

= ∫ x/(2a²) (√(x²+ a²) – √(x²– a²))dx

I = 1/(2a²) ∫x(√(x²+ a²) – √(x²– a²))dx ……(i)

Let us considered x²= t then,

On differentiating the above function we have,

d(x²) = dt

2xdx = dt

xdx = dt/2

Now on putting x²= t and xdx = dt/2 in equation (i), we get

I = 1/(2a²) ∫(√(t + a²) – √(t – a²)) dx/2

Hence, I = 1/(4a²) [2/3 (t + a²)^3/2– 2/3 (t – a²)^3/2] + c

为什么编程需要懂一点英语

问题56.∫x(TAN ^{^-1⁡x2)/(1} + X ^4)DX

解决方案：

Given that I = ∫x(tan^-1⁡x²)/(1 + x⁴) dx ……..(i)

Let us considered tan^-1x²= t then,

On differentiating the above function we have,

d(tan^-1x²) = dt

(1 × 2x)/(1 + (x²)²) dx = dt

(1 × x)/(1 + x⁴) dx = dt/2

Now on putting tan^-1⁡x²= t and x/(1 + x⁴) dx = dt/2 in equation (i), we get

I = ∫ t dx/2

= 1/2 ∫tdt

= 1/2 × t²/2 + c

= t²/4 + c – 1

= (tan^-1x²)²/4 + c

Hence, I = 1/4 (tan^-1⁡x²)²+ c

为什么编程需要懂一点英语

问题57.∫(sin ^-1 x) ³ /√(1 – x ² )dx

解决方案：

Given that I = ∫(sin^-1x)³/√(1 – x²) dx ……(i)

Let us considered sin^-1x = t then,

On differentiating the above function we have,

d(sin^-1⁡x) = dt

1/√(1 – x²) dx = dt

Now on putting sin^-1⁡x = t and 1/√(1 – x²) dx = dt in equation (i), we get

I = ∫t³ dt

= t⁴/4 + c

Hence, I = 1/4 (sin^-1x)⁴+ c

为什么编程需要懂一点英语

问题58.∫(sin⁡(2 +3log⁡x))/ x dx

解决方案：

Given that I = ∫(sin⁡(2 + 3log⁡x))/x dx ……..(i)

Let us considered 2 + 3log⁡x = t then,

On differentiating the above function we have,

d(2 + 3log⁡x) = dt

3 1/x dx = dt

dx/x = dt/3

Now on putting 2 + 3log⁡x = t and dx/x = dt/3 in equation (i), we get

I = ∫sin⁡t dt/3

= 1/3(-cos⁡t) + ct

= -1/3 cos⁡(2 + 3log⁡x) + c

Hence, I = -1/3 cos⁡(2 + 3log⁡x) + c

为什么编程需要懂一点英语

问题59。 $∫xe^{x^{2}} dx$

解决方案：

Given that I = $∫xe^{x^{2}} dx$ ……(i)

Let us considered x² = t then,

On differentiating the above function we have,

d(x²) = dt

2xdx = dt

xdx = dt/2

Now on putting x²= t and xdx = dt/2 in equation (i), we get

I = ∫e^tdt/2

= 1/2 e^t+c

= 1/2 $e^{x^{2}}$ + c

Hence, I = 1/2 $e^{x^{2}}$ + c

为什么编程需要懂一点英语

问题^{60.∫E2×/(1} + E ^X)DX

解决方案：

Given that I = ∫e^2x/(1 + e^x) dx …….(i)

Let us considered 1 + e^x= t then,

On differentiating the above function we have,

d(1 + e^x) = dt

e^xdx = dt

dx = dt/e^x

Now on putting 1 + e^x= t and dx = dt/e^xin equation (i), we get

I = ∫e^2x/t × dt/e^x

= ∫e^x/t dt

= ∫ (t – 1)/t dt

= ∫ (t/t – 1/t)dt

= t – log⁡|t| + c

= (1 + e^x) – log⁡|1 + e^x| + c

Hence, I = 1 + e^x– log⁡|1 + e^x| + c

为什么编程需要懂一点英语

问题61.∫(秒^{2⁡√x)/√xDX}

解决方案：

Given that I = ∫(sec²√x)/√x dx ……(i)

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

dx = 2tdt [√x = t])

Now on putting √x = t and dx = 2tdt in equation (i), we get

I = ∫ (sec²t)/t × 2tdt

= 2∫ sec²⁡tdt

= 2tan⁡t + c

= 2tan⁡√x + c

Hence, I = 2tan⁡√x + c

为什么编程需要懂一点英语

62.问题^{3∫tan2Xsec⁡2xDX}

解决方案：

Given that I = ∫tan³2x sec⁡2x dx

= tan²2xtan⁡2x sec⁡2x

= (sec²2x – 1)tan⁡2x sec⁡2x

= sec²2x.tan⁡2xsec⁡2x – tan⁡2xsec⁡2x

= ∫ sec²⁡2xtan⁡2xsec⁡2xdx – ∫tan⁡2xsec⁡2xdx

= ∫ sec²2xtan⁡2xsec⁡2xdx – (sec⁡2x)/2 + c

Let us considered sec⁡2x = t

2sec⁡2xtan⁡2xdx = dt

I = 1/2 ∫t² dt – (sec⁡2x)/2 + c

I = t³/6 – (sec⁡2x)/2 + c

Hence, I = (sec⁡2x)³/6 – (sec⁡2x)/2 + c

为什么编程需要懂一点英语

问题63.∫(x +√(x + 1))/(x + 2)dx

解决方案：

Given that I = ∫(x+√(x+1))/(x+2) dx …….(i)

Let us considered x + 1 = t² then,

On differentiating the above function we have,

d(x + 1) = d(t²)

dx = 2tdt

Now on putting x + 1 = t² and dx = 2tdt in equation (i), we get

I = ∫ (x + √(t²))/(x + 2) 2tdt

= 2∫((t²– 1) + t)/((t²– 1) + 2) × tdt [x + 1 = t²]

= 2∫(t²+ t – 1)/(t²+ 1) tdt

= 2∫ (t³+ t²– t)/(t² + 1) dt

= 2[∫ t³/(t²+ 1) dt + ∫ t²/(t²+ 1) dt – ∫ t/(t²+ 1) dt]

I = 2[∫t³/(t²+ 1) dt + ∫t²/(t²+ 1) dt – ∫t/(t²+ 1) dt] ……(ii)

Let I₁= ∫t³/(t²+ 1) dt

I₂= ∫t²/(t²+ 1) dt

and I₃= ∫t/(t²+ 1) dt

Now, I₁= ∫t³/(t²+ 1) dt

= ∫(t – t/(t²+ 1))dt

= t²/2 – 1/2 log⁡(t²+ 1)

I₁= t²/2 – 1/2 log⁡(t²+ 1) + c₁……..(iii)

Since, I₂= ∫t²/(t²+ 1) dt

= ∫ (t²+ 1 – 1)/(t²+ 1) dt

= ∫(t²+ 1)/(t²+ 1) dt – ∫1/(t²+ 1) dt

= ∫dt – ∫1/(t²+ 1) dt

I₂= t – tan^-1⁡(t²) + c₂ ………….(iv)

and,

I₃= ∫t/(t²+ 1) dt

= 1/2 log⁡(1 + t²) + c₃ ……..(v)

Using equations (ii), (iii), (iv) and (v), we get

I = 2[t²/2 – 1/2 log⁡(t²+ 1) + c₁+ t-tan^-1⁡(t²) + c₂– 1/2 log⁡(1 + t²) + c₃]

= 2[t²/2 + t-tan^-1⁡(t²) – log⁡(1 + t²) + c₁+ c₂+ c₃]

= 2[t²/2 + t – tan^-1⁡(t²) – log⁡(1 + t²) + c₄ [Putting c₁+ c₂+ c₃= c₄]

= t²+ 2t – 2tan^-1(t²) – 2log⁡(1 + t²) + 2c₄

= (x + 1) + 2√(x + 1) – 2tan^-1(√(x + 1)) – 2log⁡(1 + x + 1) + 2c₄

= (x + 1) + 2√(x + 1) – 2tan^-1⁡(√(x + 1)) – 2log⁡(x + 2) + c [Putting 2c₄= c]

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

为什么编程需要懂一点英语

问题64。 $∫5^{5^{5^{x}}} 5^{5^x} 5^x dx$

解决方案：

Given that I = $∫5^{5^{5^{x}}} 5^{5^x} 5^x dx$ …….(1)

Let us considered $5^{5^{5^{x}}}$ = t then

On differentiating the above function we have,

d( $5^{5^{5^{x}}}$ ) = dt

$5^{5^{5^{x}}}$ × $5^{5^x}$ × 5^x × (log⁡5)³ dx = dt

$5^{5^{5^{x}}} 5^{5^x} 5^x$ dx = dt/((log⁡5)^3 ))

Now on putting $5^{5^{5^{x}}}$ = t and $5^{5^{5^{x}}} 5^{5^x} 5^x$ dx = dt/((log⁡5)^3 )) in equation (i), we get

I = ∫dt/((log⁡5)³)

= 1/((log⁡5)³) ∫dt

= t/((log⁡5)^3) + c

Hence, I = $5^{5^{5^{x}}}$ /((log⁡5)³) + c)

为什么编程需要懂一点英语

问题65.∫1/(x√(x ⁴ – 1))dx

解决方案：

Given that I = ∫1/(x√(x⁴– 1)) dx ……….(i)

Let us considered x² = t then,

On differentiating the above function we have,

d(x²) = dt

2xdx = dt

dx = dt/2x

Now on putting x² = t and dx = dt/2x in equation (i), we get

I = ∫ 1/(x√(t²– 1)) × dt/2x

= 1/2 ∫ 1/(x² √(t²– 1)) dt

= 1/2 ∫ 1/(t√(t²– 1)) dt

= 1/2 sec^-1t + c

= 1/2 sec^-1x²+ c

Hence, I = 1/2 sec^-1x² + c

为什么编程需要懂一点英语

问题66.∫√(e ^x – 1)dx

解决方案：

Given that I = ∫√(e^x– 1) dx ……..(i)

Let us considered e^x– 1 = t² then,

On differentiating the above function we have,

d(e^x– 1) = dt(t²)

e^x dx = 2tdt

dx = 2t/e^xdt

dx = 2t/(t² + 1) dt [e^x– 1 = t²]

Now on putting e^x– 1 = t² and dx = 2tdt/(t² + 1) in equation (i), we get

I = ∫√(t²) × 2tdt/(t² + 1)

= 2∫(t × t)/(t² + 1) dt

= 2∫t²/(t² + 1) dt

= 2∫(t² + 1 – 1)/(t² + 1) dt

= 2∫[(t² + 1)/(t²+ 1) – 1/(t²+ 1)]dt

= 2∫dt – 2∫1/(t² + 1) dt

= 2t – 2tan^-1⁡(t) + c

= 2√(e^x– 1) – 2tan^-1(√(e^x– 1)) + c

Hence, I = 2√(e^x– 1) – 2tan^-1⁡√(e^x– 1) + c

为什么编程需要懂一点英语

问题67.∫1 /(x +1)(x ² + 2x + 2)dx

解决方案：

Given that I = ∫1/(x + 1)(x²+ 2x + 2) dx

= ∫1/(x + 1)((x + 1)²+ 1) dx

Let us considered x + 1 = tan u then, [tanu = Perpendicular/Base = (x + 1)/1]

On differentiating the above function we have,

dx = sec²u du [Hypotenuses = √(x²+ 2x + 2)]

I = ∫sec²u/tanu(tan²u + 1) du

= ∫ cosu/sinu du

= log| sinu |+c

= log| sin(x + 1)| + c [As we know, sin(x + 1) = P/H = (x + 1)/√(x²+ 2x + 2)]

Hence, I = log| x + 1/√(x²+ 2x + 2)| + c

为什么编程需要懂一点英语

问题^68.∫x5 /√(1 ^+×3)DX

解决方案：

Given that I = ∫ x⁵/(√(1 + x³)) dx …….(i)

Let us considered 1 + x³= t², then

On differentiating the above function we have,

d(1 + x³) = d(t²)

3x² dx = 2t * dt

dx = 2t dt/3x²

Now on putting 1 + x³= t² and dx = 2tdt/3x² in equation (i), we get

I = ∫ x⁵/√t² * 2t/3x²dt

= ∫x⁵/t * 2t/3x²dt

= 2/3∫x³ dt

= 2/3 ∫( t²– 1) dt

= 2/3[t³/3 – 2t/3] + c

Hence, I = 2/9(1 + x³)^3/2– 2 √(1 + x²)/3 + c

为什么编程需要懂一点英语

问题^{69.∫4×3√(5} – ^×2)DX

解决方案：

Given that I = ∫4x³ √(5 – x²) dx ……(i)

Let us considered 5 – x²= t²then,

On differentiating the above function we have,

d(5 – x²) = t²

-2xdx = 2tdt

dx = (-t)/x dt

Now on putting 5 – x²= t² and dx = (-t)/x dt in equation (i), we get

I = ∫4x³ √(t²) × (-t)/x dt

= -4∫ x² t × tdt

= -4∫(5 – t²) t² dt [5 – x²= t²]

= -4∫(5t²– t⁵)dt

= -20×t³/3 + 4 t⁵/5 + c

= (-20)/3 × t³+ 4/5 × t⁵+ c

= (-20)/3 × (5 – x²)^3/2+ 4/5 × (5 – x²)^5/2+ c

I = (-20)/3 × (5 -x²)^3/2+ 4/5 × (5 – x²)^5/2+ c

为什么编程需要懂一点英语

问题70.∫1/(√x+ x)dx

解决方案：

Given that I = ∫1/(√x + x) dx ……..(i)

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

Now on putting √x = t and 2√x dt = dx in equation (i), we get

I = ∫1/(t + t²) 2t × dt [Since √x = t and x = t²]

= ∫2t/(t(1 + t)) dt

= 2∫t/(1 + t) dt

= 2log⁡|1 + t| + c

= 2log⁡|1 + √x| + c

Hence, I = 2log⁡|1 + √x|+c

为什么编程需要懂一点英语

问题71.∫1/(x ² (x ⁴ + 1) ^3/4 )dx

解决方案：

Given that I = 1/(x² (x⁴+1)^3/4)

Multiplying and dividing by x^-3, we obtain

(x^-3/(x²x^-3 (x⁴+ 1)^3/4) = (x^-3 (x⁴+ 1)^-3/4/(x²x^-3))

= (x⁴+ 1)^-3/4/(x⁵(x⁴)^-3/4

= 1/x⁵ ((x⁴+ 1)/x⁴)^-3/4

= 1/x⁵ (1 + 1/x⁴)^-3/4

Let us considered 1/x⁴ = t

-4/x⁵dx = dt

1/x⁵dx = -dt/4

I = ∫ 1/x⁵(1 + 1/x⁴)^-3/4 dx

= -1/4 ∫(1 + t)^-3/4 dt

= -1/4 [(1 + t)^1/4)/(1/4)] + c

= -1/4(1 + 1/x⁴)^1/4/(1/4) + c

Hence, I = -(1 + 1/x⁴)^1/4+ c

为什么编程需要懂一点英语

问题^{72.∫(sin⁡5×)/(COS} ^4×)DX

解决方案：

Given that I = ∫(sin⁡⁵x)/(cos⁴⁡x) dx ……(i)

Let us considered cos⁡x = t then,

On differentiating the above function we have,

d(cos⁡x) = dt

-sin⁡xdx = dt

dx = -dt/(sin⁡x)

Now on putting cos⁡x = t and dx = -dt/(sin⁡x) in equation (i), we get

I = ∫(sin⁵⁡x)/t⁴× -dt/(sin⁡x)

= -∫(sin⁴⁡x)/t⁴ dt

= -∫(1 – cos²x)²/t⁴dt

= -∫(1 – t²)²/t⁴dt

= -∫(1 + t⁴– 2t²)/t⁴dt

= -∫(1/t⁴ + t⁴/t⁴ – (2t²)/t⁴)dt

= -∫(t^-4+ 1 – 2t^-2)dt

= -[t^-3/(-3) + t – 2 t^-1/(-1)] + c

= 1/3 * 1/t³– t – 1/t + c

Hence, I = 1/3 * 1/cos³x – cosx -2/cosx + c

为什么编程需要懂一点英语

问题49。

问题50。

问题51.∫(cos⁡√x)/√xdx

问题52.∫sin(tan -1 x)/(1 + x 2 )dx

问题53.∫(sin⁡(log⁡x))/ x dx

问题54。

问题55.∫x/(√(x 2 + a 2 )+√(x 2 – a 2 ))dx

问题56.∫x(TAN -1⁡x2)/(1 + X 4)DX

问题57.∫(sin -1 x) 3 /√(1 – x 2 )dx

问题58.∫(sin⁡(2 +3log⁡x))/ x dx

问题59。

问题60.∫E2×/(1 + E X)DX

问题61.∫(秒2⁡√x)/√xDX

62.问题3∫tan2Xsec⁡2xDX

问题63.∫(x +√(x + 1))/(x + 2)dx

问题64。

问题65.∫1/(x√(x 4 – 1))dx

问题66.∫√(e x – 1)dx

问题67.∫1 /(x +1)(x 2 + 2x + 2)dx

问题68.∫x5 /√(1 +×3)DX

问题69.∫4×3√(5 – ×2)DX

问题70.∫1/(√x+ x)dx

问题71.∫1/(x 2 (x 4 + 1) 3/4 )dx

问题72.∫(sin⁡5×)/(COS 4×)DX

问题49。 $∫5^{x + tan^{-1}x}(\frac{x^2 + 2}{x^2 + 1})dx$

问题50。 $∫\frac{e^{msin^{-1}x}}{\sqrt{1-x^2}} dx$

问题52.∫sin(tan ^-1 x)/(1 + x ² )dx

问题54。 $∫\frac{e^{mtan^{-1}x}}{(1+x^2)}dx$

问题55.∫x/(√(x ² + a ² )+√(x ² – a ² ))dx

问题56.∫x(TAN ^{^-1⁡x2)/(1} + X ^4)DX

问题57.∫(sin ^-1 x) ³ /√(1 – x ² )dx

问题59。 $∫xe^{x^{2}} dx$

问题^{60.∫E2×/(1} + E ^X)DX

问题61.∫(秒^{2⁡√x)/√xDX}

62.问题^{3∫tan2Xsec⁡2xDX}

问题64。 $∫5^{5^{5^{x}}} 5^{5^x} 5^x dx$

问题65.∫1/(x√(x ⁴ – 1))dx

问题66.∫√(e ^x – 1)dx

问题67.∫1 /(x +1)(x ² + 2x + 2)dx

问题^68.∫x5 /√(1 ^+×3)DX

问题^{69.∫4×3√(5} – ^×2)DX

问题71.∫1/(x ² (x ⁴ + 1) ^3/4 )dx

问题^{72.∫(sin⁡5×)/(COS} ^4×)DX