第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.1 |设置 1

计算以下定积分：

问题 1。 $\int_{4}^{9} \frac{1}{\sqrt{x}}dx$

解决方案：

We have,

I = $\int_{4}^{9} \frac{1}{\sqrt{x}}dx$

I = $\left[\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right]^9_4$

I = $\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]^9_4$

I = $\left[2\sqrt{x}\right]^9_4$

I = 2[√9 – √4 ] 

I = 2 (3 − 2)

I = 2 (1)

I = 2

Therefore, the value of $\int_{4}^{9} \frac{1}{\sqrt{x}}dx$is 2.

问题2。 $\int_{-2}^{3} \frac{1}{x+7}dx$

解决方案：

We have,

I = $\int_{-2}^{3} \frac{1}{x+7}dx$

I = $\left[log(x+7)\right]^3_{-2}$

I = log (3 + 7) − log (−2 + 7)

I = log 10 − log 5

I = $log\frac{10}{5}$

I = log 2

Therefore, the value of $\int_{-2}^{3} \frac{1}{x+7}dx$is log 2.

问题 3。 $\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx$

Let x = sin t, so we have, 

=> dx = cos t dt

Now, the lower limit is,

=> x = 0 

=> sin t = 0

=> t = 0

Also, the upper limit is,

=> x = 1/2

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{1-sin^2t}}costdt$

I = $\int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{cos^2t}}costdt$

I = $\int_{0}^{\frac{\pi}{6}} (\frac{1}{cost})costdt$

I = $\int_{0}^{\frac{\pi}{6}} 1dt$

I = $\left[t\right]_0^{\frac{\pi}{6}}$

I =  π/6 – 0

I = π/6​

Therefore, the value of $\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx$is π/6.

问题 4。 $\int_{0}^{1} \frac{1}{1+x^2}dx$

解决方案：

We have,

I = $\int_{0}^{1} \frac{1}{1+x^2}dx$

I = $\left[tan^{-1}x\right]_0^1$

I = $tan^{-1}1-tan^{-1}0$

I = $\frac{\pi}{4}-0$

I = π/4

Therefore, the value of $\int_{0}^{1} \frac{1}{1+x^2}dx$is π/4.

问题 5。 $\int_{2}^{3} \frac{x}{x^2+1}dx$

解决方案：

We have,

I = $\int_{2}^{3} \frac{x}{x^2+1}dx$

Let x² + 1 = t, so we have,

=> 2x dx = dt

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x² + 1

=> t = (2)² + 1

=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x² + 1

=> t = (3)² + 1

=> t = 9 + 1

=> t = 10

So, the equation becomes,

I = $\int_{5}^{10} \frac{1}{2t}dt$

I = $\frac{1}{2}\int_{5}^{10} \frac{1}{t}dt$

I = $\frac{1}{2}\left[logt\right]^{10}_5$

I = 1/2[log10 – log5] 

I = 1/2[log10/5]

I = 1/2[log2]

I = log√2

Therefore, the value of $\int_{2}^{3} \frac{x}{x^2+1}dx$is log√2.

问题 6。 $\int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx$

解决方案：

We have,

I = $\int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx$

I = $\int_{0}^{\infty} \frac{1}{b^2}\left(\frac{1}{\frac{a^2}{b^2}+x^2}\right)dx$

I = $\frac{1}{b^2}\int_{0}^{\infty}\frac{1}{\frac{a^2}{b^2}+x^2}dx$

I = $\frac{1}{b^2}\left[\frac{b}{a}tan^{-1}\frac{bx}{a}\right]^{\infty}_{0}$

I = $\frac{1}{ab}\left[tan^{-1}\frac{bx}{a}\right]^{\infty}_{0}$

I = 1/ab[tan^-1∞ – tan^-10] 

I = 1/ab[π/2 – 0]  

I = 1/ab[π/2]  

I = π/2ab   

Therefore, the value of $\int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx$is π/2ab.

问题 7。 $\int_{-1}^{1} \frac{1}{1+x^2}dx$

解决方案：

We have,

I = $\int_{-1}^{1} \frac{1}{1+x^2}dx$

I = $\left[tan^{-1}x\right]_{-1}^{1}$

I = [tan^-11 – tan^-1(-1)]  

I = [π/4 – (-π/4)]   

I = [π/4 + π/4]  

I = 2π/4 

I = π/2  

Therefore, the value of $\int_{-1}^{1} \frac{1}{1+x^2}dx$is π/2.

问题 8。 $\int_{0}^{\infty} e^{-x}dx$

解决方案：

We have,

I = $\int_{0}^{\infty} e^{-x}dx$

I = $\left[-e^{-x}\right]^{\infty}_{0}$

I = -e^–∞ – (-e⁰) 

I = − 0 + 1

I = 1

Therefore, the value of $\int_{0}^{\infty} e^{-x}dx$is 1.

问题 9。 $\int_{0}^{1} \frac{x}{x+1}dx$

解决方案：

We have,

I = $\int_{0}^{1} \frac{x}{x+1}dx$

I = $\int_{0}^{1} \frac{(x+1)-1}{x+1}dx$

I = $\int_{0}^{1} \frac{x+1}{x+1}dx-\int_{0}^{1}\frac{1}{x+1}dx$

I = $\int_{0}^{1} 1dx-\int_{0}^{1}\frac{1}{x+1}dx$

I = $\left[x\right]^1_0-\left[log(x+1)\right]^1_0$

I = [1 − 0] − [log(1 + 1) − log(0 + 1)]

I = 1 − [log2 − log1]

I = 1 – log2/1 

I = 1 − log 2

I = log e − log 2

I = loge/2 

Therefore, the value of $\int_{0}^{1} \frac{x}{x+1}dx$ is loge/2.

问题 10。 $\int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx$

I = $\int_{0}^{\frac{\pi}{2}} (sinx)dx+\int_{0}^{\frac{\pi}{2}}(cosx)dx$

I = $\left[-cosx\right]_{0}^{\frac{\pi}{2}}+\left[sinx\right]_{0}^{\frac{\pi}{2}}$

I = [-cosπ/2 + cos0] + [sinπ/2 – sin0] 

I = [−0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx$is 2.

问题 11。 $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx$

解决方案：

We have,

I = $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx$

I = $\left[log(sinx)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

I = log(sinπ/2) – log(sinπ/4)

I = log1 – log1/√2 

I = $log\frac{1}{\frac{1}{\sqrt{2}}}$

I = log√2  

Therefore, the value of $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx$is log√2.

问题 12。 $\int_{0}^{\frac{\pi}{4}} secxdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}} secxdx$

I = $\left[log(secx+tanx)\right]^{\frac{\pi}{4}}_0$

I = log(secπ/4 + tanπ/4 – log(sec0 + tan0) 

I = log(√2 + 1) – log(1 + 0) 

I = $log(\frac{\sqrt{2}+1}{1})$

I = log(√2 + 1) 

Therefore, the value of $\int_{0}^{\frac{\pi}{4}} secxdx$is log(√2 + 1).

问题 13。 $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx$

解决方案：

We have,

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx$

I = $\left[log|cosecx-cotx|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

I = [log|cosecπ/4 – cotπ/4|] – [log|cosecπ/6 – cotπ/6|]

I = [log|√2 – 1|] – [log|2 – √3|] 

I = $log(\frac{\sqrt{2}-1}{2-\sqrt{3}})$

Therefore, the value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx$is $log(\frac{\sqrt{2}-1}{2-\sqrt{3}})$.

问题 14。 $\int_{0}^{1} \frac{1-x}{1+x}dx$

解决方案：

We have, 

I = $\int_{0}^{1} \frac{1-x}{1+x}dx$

Let x = cos 2t, so we have,

=> dx = –2 sin 2t dt

Now, the lower limit is,

=> x = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is,

=> x = 1

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = $\int_{\frac{\pi}{4}}^{0} \frac{1-cos2t}{1+cos2t}(-2sin2t)dt$

I = $\int_{\frac{\pi}{4}}^{0}\frac{2sin^2t}{2cos^2t}(-2sin2t)dt$

I = $\int^{\frac{\pi}{4}}_{0}\frac{sin^2t}{cos^2t}(2sin2t)dt$

I = $\int^{\frac{\pi}{4}}_{0}\frac{sin^2t}{cos^2t}(4sintcost)dt$

I = $\int^{\frac{\pi}{4}}_{0}\frac{4sin^3t}{cost}dt$

Let cos t = z, so we have,

=> – sin t dt = dz

=> sin t dt = – dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = π/4

=> z = cos t

=> z = cos π/4

=> z = 1/√2

So, the equation becomes,

I = $\int^{\frac{\pi}{4}}_{0}\frac{4sin^2t(sint)}{cost}dt$

I = $\int^{\frac{1}{\sqrt{2}}}_{1}\frac{-4(1-z^2)}{z}dz$

I = $-4\int^{\frac{1}{\sqrt{2}}}_{1}\frac{1-z^2}{z}dz$

I = $-4\int^{\frac{1}{\sqrt{2}}}_{1}(\frac{1}{z}-\frac{z^2}{z})dz$

I = $-4\int^{\frac{1}{\sqrt{2}}}_{1}(\frac{1}{z}-z)dz$

I = $-4\left[logz-\frac{z^2}{2}\right]^{\frac{1}{\sqrt{2}}}_1$

I = -4[(log1/√2 – 1/2(2)) – (log1 – 1/2)]

I = -4[(log1/√2 – 1/4) – (0 – 1/2)]

I = -4[log1/√2 – 1/4 – 0 + 1/2]

I = -4[-log√2 + 1/4]

I = 4log√2 – 1 

I = 4 × 1/2log2 – 1

I = 2log2 – 1 

Therefore, the value of $\int_{0}^{1} \frac{1-x}{1+x}dx$is 2log2 – 1.

问题 15。 $\int_{0}^{\pi} \frac{1}{1+sinx}dx$

解决方案：

We have,

I = $\int_{0}^{\pi} \frac{1}{1+sinx}dx$

I = $\int_{0}^{\pi} \frac{1-sinx}{(1+sinx)(1-sinx)}dx$

I = $\int_{0}^{\pi} \frac{1-sinx}{1-sin^2x}dx$

I = $\int_{0}^{\pi} \frac{1-sinx}{cos^2x}dx$

I = $\int_{0}^{\pi}\frac{1}{cos^2x}-\frac{sinx}{cos^2x}dx$

I = $\int_{0}^{\pi}sec^2x-\frac{sinx}{cosx(cosx)}dx$

I = $\int_{0}^{\pi}(sec^2x-tanxsecx)dx$

I = $\int_{0}^{\pi}sec^2xdx-\int_{0}^{\pi}tanxsecxdx$

I = $\left[tanx\right]^{\pi}_0-\left[secx\right]^{\pi}_0$

I = [tan π – tan0] – [sec π – sec 0]

I = [0 – 0] – [–1 – 1]

I = 0 – (–2)

I = 2

Therefore, the value of $\int_{0}^{\pi} \frac{1}{1+sinx}dx$is 2.

问题 16。 $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx$

解决方案：

We have,

I = $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{(1+sinx)(1-sinx)}dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{1-sin^2x}dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{cos^2x}dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{1}{cos^2x}-\frac{sinx}{cos^2x}dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}sec^2x-\frac{sinx}{cosx(cosx)}dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{-\pi}{4}}(sec^2x-tanxsecx)dx$

I = $\int_{\frac{-\pi}{4}}^{\frac{-\pi}{4}}sec^2xdx-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}tanxsecxdx$

I = $\left[tanx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}-\left[secx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}$

I = $\left[tanx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}-\left[secx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}$

I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]

I = [1 – (–1)] – [sec π/4 – sec (π/4)]

I = 2 – 0

I = 2

Therefore, the value of $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx$is 2.

问题 17。 $\int_{0}^{\frac{\pi}{2}} cos^2xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}} cos^2xdx$

I = $\int_{0}^{\frac{\pi}{2}} (\frac{1+cos2x}{2})dx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}} (1+cos2x)dx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}1dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos2xdx$

I = $\frac{1}{2}\left[x\right]^{\frac{\pi}{2}}_0+\frac{1}{2}\left[\frac{sin2x}{2}\right]^{\frac{\pi}{2}}_0$

I = $\frac{1}{2}\left[x\right]^{\frac{\pi}{2}}_0+\frac{1}{4}\left[sin2x\right]^{\frac{\pi}{2}}_0$

I = 1/2[π/2 – 0] + 1/4[sinπ – sin0]

I = 1/2[π/2] + 1/4[0 – 0]

I = π/4 

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} cos^2xdx$ is π/4.

问题 18。 $\int_{0}^{\frac{\pi}{2}} cos^3xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}} cos^3xdx$

I = $\int_{0}^{\frac{\pi}{2}} \frac{cos3x+3cosx}{4}dx$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}} (cos3x+3cosx)dx$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}} cos3xdx+\frac{3}{4}\int_{0}^{\frac{\pi}{2}}cosxdx$

I = $\frac{1}{4}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{2}}_0+\frac{3}{4}[sinx]^{\frac{\pi}{2}}_{0}$

I = $\frac{1}{12}\left[sin3x\right]^{\frac{\pi}{2}}_0+\frac{3}{4}[sinx]^{\frac{\pi}{2}}_{0}$

I = 1/12 [-1 – 0] + 3/4[1 – 0]

I = 3/4 – 1/12

I = (9 – 1)/12

I = 8/12

I = 2/3

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} cos^3xdx$ is 2/3.

问题 19。 $\int_{0}^{\frac{\pi}{6}} cosxcos2xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{6}} cosxcos2xdx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{6}} 2cosxcos2xdx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{6}} (cos3x + cosx)dx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{6}}cos3xdx + \frac{1}{2}\int_{0}^{\frac{\pi}{6}}cosxdx$

I = $\frac{1}{2}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{6}}_0 + \frac{1}{2}\left[sinx\right]^{\frac{\pi}{6}}_0$

I = $\frac{1}{6}\left[sin3x\right]^{\frac{\pi}{6}}_0 + \frac{1}{2}\left[sinx\right]^{\frac{\pi}{6}}_0$

I = 1/6[sinπ/2 – sin0] + 1/2[sinπ/6 – sin0]

I = 1/6[1 – 0] + 1/2[1/2 – 0]

I = 1/6 + 1/4

I = (4 + 6)/24 

I = 10/24 

I = 5/12 

Therefore, the value of $\int_{0}^{\frac{\pi}{6}} cosxcos2xdx$ is 5/12.

问题 20。 $\int_{0}^{\frac{\pi}{2}} sinxsin2xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}} sinxsin2xdx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}} 2sinxsin2xdx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}} (cosx - cos3x)dx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cosxdx - \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos3xdx$

I = $\frac{1}{2}\left[sinx\right]^{\frac{\pi}{2}}_0-\frac{1}{2}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{2}}_0$

I = $\frac{1}{2}\left[sinx\right]^{\frac{\pi}{2}}_0-\frac{1}{6}[sin3x]^{\frac{\pi}{2}}_0$

I = 1/2[sinπ/2 – sin0] – 1/6[sin3π/2 – sin0]

I = 1/2[1 – 0] – 1/6[-1 – 0]

I = 1/2 – 1/6(-1)

I = 1/2 + 1/6 

I = (6 + 2)/12

I = 8/12 

I = 2/3

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} sinxsin2xdx$ is 2/3.

问题 21。 $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx$

解决方案：

We have,

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{sinx}{cosx}+\frac{cosx}{sinx})^2dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{sin^2x+cos^2x}{cosxsinx})^2dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{1}{cosxsinx})^2dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{1}{cos^2xsin^2x}dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{4}{4cos^2xsin^2x}dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{4}{(sin2x)^2}dx$

I = $4\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}cosec^22xdx$

I = $4\left[\frac{-cot2x}{2}\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}$

I = $2\left[-cot2x\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}$

I = 2[-cotπ/2 + cot2π/3]

I = 2[-1/√3 – 0]

I = -2/√3 

Therefore, the value of $\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx$ is -2/√3.

问题 22。 $\int_{0}^{\frac{\pi}{2}} cos^4xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}} cos^4xdx$

I = $\int_{0}^{\frac{\pi}{2}} (cos^2x)^2dx$

I = $\int_{0}^{\frac{\pi}{2}} (\frac{1+cos2x}{2})^2dx$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+cos2x)^2dx$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+cos^22x+2cos2x)dx$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+\frac{1+cos4x}{2}+2cos2x)dx$

I = $\frac{1}{4}\left[x+\frac{x}{2}+\frac{sin4x}{8}+sin2x\right]^{\frac{\pi}{2}}_0$

I = 1/4[π/2 + π/4 + 0 + 0 – 0 – 0 – 0 – 0]

I = 1/4[3π/4] 

I = 3π/16 

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} cos^4xdx$is 3π/16.