第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.1 |设置 3

计算以下定积分：

问题 45。 $\int_{1}^{4} \frac{x^2+x}{\sqrt{2x+1}}dx$

解决方案：

We have,

I = $\int_{1}^{4} \frac{x^2+x}{\sqrt{2x+1}}dx$

Let 2x + 1 = t², so we have,

=> 2 dx = 2t dt

=> dx = t dt

Now, the lower limit is, x = 1

=> t² = 2x + 1

=> t² = 2(1) + 1

=> t² = 3

=> t = √3 

Also, the upper limit is, x = 4

=> t² = 2x + 1

=> t² = 2(4) + 1

=> t² = 9

=> t = 3 

So, the equation becomes,

I = $\int_{\sqrt{3}}^{3} \frac{(\frac{t^2-1}{2})^2+(\frac{t^2-1}{2})}{t}tdt$

I = $\int_{\sqrt{3}}^{3} (\frac{t^2-1}{2})^2+(\frac{t^2-1}{2})dt$

I = $\frac{1}{4}\int_{\sqrt{3}}^{3} (t^4+1-2t^2+2t^2-2)dt$

I = $\frac{1}{4}\int_{\sqrt{3}}^{3} (t^4-1)dt$

I = $\frac{1}{4}\left[\frac{t^5}{5}-t\right]^3_{\sqrt{3}}$

I = 1/4 [3⁵/5 – 3 – (√3)⁵/5 + √3]

I = 1/4[243/5 – 3 – 9√3/5 + √3]

I = 1/4((243 – 15 – 9√3 + 5√3)/5)

I = 1/4[(228 – 4√3)/5]

I = 1/4[4(57 – √3)/5]

I = (57 – √3)/5 

Therefore, the value of $\int_{1}^{4} \frac{x^2+x}{\sqrt{2x+1}}dx$is (57 – √3)/5.

问题 46。 $\int_{0}^{1} x(1-x)^5dx$

解决方案：

We have,

I = $\int_{0}^{1} x(1-x)^5dx$

By using binomial theorem in the expansion of (1 – x)⁵, we get,

I = $\int_{0}^{1} x[1^5+^5C_1(-x)+^5C_2(-x)^2+^5C_3(-x)^3+^5C_4(-x)^4+^5C_5(-x)^5]dx$

I = $\int_{0}^{1} x(1-5x+10x^2-10x^3+5x^4-x^5)dx$

I = $\int_{0}^{1} (x-5x^2+10x^3-10x^4+5x^5-x^6)dx$

I = $\left[\frac{x^2}{2}-\frac{5x^3}{3}+\frac{10x^4}{4}-\frac{10x^5}{5}+\frac{5x^6}{6}-\frac{x^7}{7}\right]^{1}_0$

I = 1/2 – 5/3 + 10/4 – 10/5 + 5/6 – 1/7

I = 1/2 – 5/3 + 5/3 – 2 + 5/6 – 1/7

I = 1/2 – 2 + 5/6 – 1/7 

I = 1/42

Therefore, the value of $\int_{0}^{1} x(1-x)^5dx$ is 1/42.

问题 47。 $\int_{1}^{2} (\frac{x-1}{x^2})e^xdx$

解决方案：

We have,

I = $\int_{1}^{2} (\frac{x-1}{x^2})e^xdx$

I = $\int_{1}^{2} \frac{xe^x-e^x}{x^2}dx$

I = $\int_{1}^{2} (\frac{e^x}{x}-\frac{e^x}{x^2})dx$

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

By using integration by parts, we get,

I = $\frac{e^x}{x}-\int ((\frac{-1}{x^2})\int e^xdx)dx-\int \frac{e^x}{x^2}dx$

I = $\frac{e^x}{x}+\int\frac{e^x}{x^2}dx-\int \frac{e^x}{x^2}dx$

I = e^x/x

So we get,

I = $\left[\frac{e^x}{x}\right]^2_1$

I = e²/2 – e¹/1

I = e²/2 – e

Therefore, the value of $\int_{1}^{2} (\frac{x-1}{x^2})e^xdx$is e²/2 – e.

问题 48。 $\int_{0}^{1} (xe^{2x}+sin\frac{\pi x}{2})dx$

解决方案：

We have,

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

By using integration by parts in first integral, we get,

I = $\frac{xe^{2x}}{2}-\frac{1}{2}\int e^{2x}dx+\left[\frac{-cos\frac{\pi x}{2}}{\frac{\pi}{2}}\right]^1_0$

I = xe^2x/2 – (1/2)(e^2x/2) + 2/π[1 – 0]

I = xe^2x/2 – e^2x/4 + 2/π

So we get,

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

I = [e²/2 + e²/4 – 0 + 1/4] + 2/π

I = e²/4 + 1/4 + 2/π

Therefore, the value of $\int_{0}^{1} (xe^{2x}+sin\frac{\pi x}{2})dx$ is e²/4 + 1/4 + 2/π.

问题 49。 $\int_{0}^{1} (xe^{x}+cos\frac{\pi x}{4})dx$

解决方案：

We have,

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

By using integration by parts in first integral, we get,

I = $xe^{x}-\int ((1)\int e^xdx)dx+ \left[\frac{sin\frac{\pi x}{4}}{\frac{\pi x}{4}}\right]^1_0$

I = $xe^{x}-\int e^xdx+\frac{4}{\pi}[\frac{1}{\sqrt{2}}-0]$

I = $xe^{x}-e^x+\frac{4}{\pi}[\frac{1}{\sqrt{2}}]$

I = $xe^{x}-e^x+\frac{2\sqrt{2}}{\pi}$

So we get,

I = $\left[xe^{x}-e^x\right]^1_0+\frac{2\sqrt{2}}{\pi}$

I = $\left[e^{x}(x-1)\right]^1_0+\frac{2\sqrt{2}}{\pi}$

I = [e¹(1 – 1) – e⁰(0 – 1)] + 2√2/π

I = [0 – (-1)] + 2√2/π

I = 1 + 2√2/π

Therefore, the value of $\int_{0}^{1} (xe^{x}+cos\frac{\pi x}{4})dx$ is 1 + 2√2/π.

问题 50。 $\int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-sinx}{1-cosx})dx$

解决方案：

We have,

I = $\int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-sinx}{1-cosx})dx$

I = $\int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-2sin\frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}})dx$

I = $-\int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{-1}{2}cosec^2\frac{x}{2}+cot\frac{x}{2})dx$

I = $\left[-e^xcot\frac{x}{2}\right]^\pi_\frac{\pi}{2}$

I = -e^π cotπ/2 + e^π/2 cotπ/4

I = 0 + e^π/2(1)

I = e^π/2

Therefore, the value of $\int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-sinx}{1-cosx})dx$is e^π/2.

问题 51。 $\int_{0}^{2\pi} e^{\frac{x}{2}}\sin(\frac{x}{2}+\frac{\pi}{4})dx$

解决方案：

We have,

I = $\int_{0}^{2\pi} e^{\frac{x}{2}}\sin(\frac{x}{2}+\frac{\pi}{4})dx$

I = $\int_{0}^{2\pi} e^{\frac{x}{2}}(sin\frac{x}{2}cos\frac{\pi}{4}+cos\frac{x}{2}sin\frac{\pi}{4})dx$

I = $\int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}sin\frac{x}{2}+\frac{1}{\sqrt{2}}cos\frac{x}{2})dx$

I = $\int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}sin\frac{x}{2})dx+\int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx$

By using integration by parts in first integral, we get,

I = $\frac{1}{\sqrt{2}}\left[sin\frac{x}{2}\int_{0}^{2\pi} e^{\frac{x}{2}}dx-\int (\int (e^{\frac{x}{2}}dx)\frac{sin\frac{x}{2}}{2})dx\right]+\int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx$

I = $\frac{1}{\sqrt{2}}\left[sin\frac{x}{2}(2e^{\frac{x}{2}})\right]^{2\pi}_0-\int_{0}^{2\pi}e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx+\int_{0}^{2\pi}e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx$

I = $\frac{1}{\sqrt{2}}\left[sin\frac{x}{2}(2e^{\frac{x}{2}})\right]^{2\pi}_0$

I = 1/√2[sinπ(2e^π) – 0]

I = 1/√2[0 – 0]

I = 0

Therefore, the value of $\int_{0}^{2\pi} e^{\frac{x}{2}}\sin(\frac{x}{2}+\frac{\pi}{4})dx$is 0.

问题 52。 $\int_{0}^{2\pi} e^x\cos(\frac{x}{2}+\frac{\pi}{4})dx$

解决方案：

We have,

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

By using integration by parts, we get,

I = e^xcos(x/2 + π/4) + 1/2∫e^xsin(x/2 + π/4)

I = e^x cos(x/2 + π/4) + 1/2[ e^xsin(x/2 + π/4) – 1/2 ∫e^xcos(x/2 + π/4)dx]

I = e^xcos(x/2 + π/4) + 1/2e^xsin(x/2 + π/4) – 1/4I

$\frac{5I}{4}=\left[e^x\cos(\frac{x}{2}+\frac{\pi}{4})+\frac{1}{2}e^xsin(\frac{x}{2}+\frac{\pi}{4})\right]^{2\pi}_0$

$\frac{5I}{4}=\left[\frac{-1}{\sqrt{2}}(e^{2\pi}+1)-\frac{1}{2\sqrt{2}}(e^{2\pi}+1)\right]$

5I/4 = -3/ 2√2(e^2π + 1)

I = -3√2/5(e^2π + 1)

Therefore, the value of $\int_{0}^{2\pi} e^x\cos(\frac{x}{2}+\frac{\pi}{4})dx$is -3√2/5(e^2π + 1).

问题 53。 $\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$

解决方案：

We have,

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

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

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

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

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

I = $\left[\frac{2}{3}(1+x)^{\frac{3}{2}}\right]^1_0+\left[\frac{2}{3}x^{\frac{3}{2}}\right]^1_0$

I = 2/3[2^3/2 – 1] + 2/3[1 – 0]

I = $(\frac{2^{1+\frac{3}{2}}}{3})-\frac{2}{3}+\frac{2}{3}$

I = 2^5/2/3 

Therefore, the value of $\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$is 2^5/2/3.

问题 54。 $\int_{1}^{2} \frac{x}{(x+1)(x+2)}dx$

解决方案：

We have,

I = $\int_{1}^{2} \frac{x}{(x+1)(x+2)}dx$

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

I = $-\left[log(x+1)\right]^2_1+\left[2log(x+2)\right]^2_1$

I = $-\left[log(x+1)\right]^2_1+2\left[log(x+2)\right]^2_1$

I = -log3 + log2 + 2[log4 – log3]

I = -log3 + log2 + 2[2log2 – log3]

I = -log3 + log2 + 4log2 – 2log3

I = 5log2 – 3log3

I = log2⁵ – log3³

I = log32 – log27

I = log32/27 

Therefore, the value of $\int_{1}^{2} \frac{x}{(x+1)(x+2)}dx$is log32/27.

问题 55。 $\int_{0}^{\frac{\pi}{2}} sin^3xdx$

解决方案：

We have,

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

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

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

Let cos x = t, so we have,

=> – sin x dx = dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x =  π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I = $\int_{1}^{0} (t^2-1)dt$

I = $\int_{1}^{0}t^2dt-\int_{1}^{0}(1)dt$

I = $\left[\frac{t^2}{3}\right]^{0}_1-\left[t\right]^0_1$

I = [0 – 1/3] – [0 – 1]

I = [-1/3] – [-1]

I = -1/3 + 1

I = 2/3

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

问题 56。 $\int_{0}^{\pi} (sin^2\frac{x}{2}-cos^2\frac{x}{2})dx$

解决方案：

We have,

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

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

I = $-\int_{0}^{\pi}cosxdx$

I = $\left[-sinx\right]^\pi_0$

I = -sinπ + sin0 

I = 0

Therefore, the value of $\int_{0}^{\pi} (sin^2\frac{x}{2}-cos^2\frac{x}{2})dx$ is 0.

问题 57。 $\int_{1}^{2}(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$

解决方案：

 We have,

I = $\int_{1}^{2}(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$

Let 2x = t, so we have,

=> 2x dx = dt

Now, the lower limit is, x = 1

=> t = 2x

=> t = 2(1)

=> t = 2

Also, the upper limit is, x =  2

=> t = 2x

=> t = 2(2)

=> t = 4

So, the equation becomes,

I = $\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{4}{2t^2})e^{t}dt$

I = $\int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^{t}dt$

I = $\int_{2}^{4}\frac{e^t}{t}dt-\int_{2}^{4}\frac{e^t}{t^2}dt$

By using integration by parts in first integral, we get,

I = $\left[\frac{e^t}{t}\right]^4_2-\int (\frac{-1}{t^2}\int e^tdx)dx-\int_{2}^{4}\frac{e^t}{t^2}dt$

I = $\left[\frac{e^t}{t}\right]^4_2+\int_{2}^{4}\frac{e^t}{t^2}dt-\int_{2}^{4}\frac{e^t}{t^2}dt$

I = $\left[\frac{e^t}{t}\right]^4_2$

I = e⁴/4 – e²/2

Therefore, the value of $\int_{1}^{2}(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$ is e⁴/4 – e²/2.

问题 58。 $\int_{1}^{2}\frac{1}{\sqrt{(x-1)(2-x)}}dx$

解决方案：

We have,

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

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

I = $\int_{1}^{2}\frac{1}{\sqrt{-x^2+3x-2}}dx$

I = $\int_{1}^{2}\frac{1}{\sqrt{-(x-\frac{3}{2})^2+\frac{1}{4}}}dx$

I = $\int_{1}^{2}\frac{1}{\sqrt{(\frac{1}{2})^2-(x-\frac{3}{2})^2}}dx$

I = $\left[sin^{-1}(2x-3)\right]_{1}^{2}$

I = [sin^-1(1) – sin^-1(-1)]

I = π/2 – (-π/2)

I = π/2 + π/2

I =  π

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

问题 59. 如果$\int_{0}^{k}\frac{1}{2+8x^2}dx=\frac{\pi}{16}$ ，求k的值。

解决方案：

We have,

=> $\int_{0}^{k}\frac{1}{2+8x^2}dx=\frac{\pi}{16}$

=> $\int_{0}^{k}\frac{1}{2(1+4x^2)}dx=\frac{\pi}{16}$

=> $\int_{0}^{k}\frac{1}{2(1+(2x)^2)}dx=\frac{\pi}{16}$

=> $\left[\frac{tan^{-1}2x}{4}\right]_{0}^{k}=\frac{\pi}{16}$

=> tan^-12k/4 – tan^-10 = π/16

=> tan^-12k/4 – 0 = π/16

=> tan^-12k/4 = π/16

=> tan^-12k = π/4

=> 2k = tanπ/4

=> 2k = 1

=> k = 1/2

Therefore, the value of k is 1/2.

问题 60. 如果$\int_{0}^{a}3x^2dx=8$ ，求k的值。

解决方案：

We have,

=> $\int_{0}^{a}3x^2dx=8$

=> $\left[\frac{3(x^{2+1})}{2+1}\right]^a_0=8$

=> $\left[\frac{3(x^{3})}{3}\right]^a_0=8$

=> $\left[x^3\right]^a_0=8$

=> a³ – 0 = 8

=> a³ = 8

=> a = 2

Therefore, the value of a is 2.

问题 61。 $\int_\pi^\frac{3\pi}{2}\sqrt{1-cos2x}dx$

解决方案：

We have,

I = $\int_\pi^\frac{3\pi}{2}\sqrt{1-cos2x}dx$

I = $\int_\pi^\frac{3\pi}{2}\sqrt{2sin^2x}dx$

I = $\int_\pi^\frac{3\pi}{2}\sqrt{2}sinxdx$

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

I = -[√2cos3π/2 – √2cosπ]

I = -(-√2 – 0)

I = √2

Therefore, the value of $\int_\pi^\frac{3\pi}{2}\sqrt{1-cos2x}dx$ is √2.

问题 62。 $\int_0^{2\pi}\sqrt{1+sin\frac{x}{2}}dx$

解决方案：

We have,

I = $\int_0^{2\pi}\sqrt{1+sin\frac{x}{2}}dx$

I = $\int_0^{2\pi}\sqrt{sin^2\frac{x}{4}+cos^2\frac{x}{4}+2sin\frac{x}{4}cos\frac{x}{4}}dx$

I = $\int_0^{2\pi}\sqrt{(sin\frac{x}{4}+cos\frac{x}{4})^2}dx$

I = $\int_0^{2\pi}(sin\frac{x}{4}+cos\frac{x}{4})dx$

I = $\int_0^{2\pi}sin\frac{x}{4}dx+\int_0^{2\pi}cos\frac{x}{4}dx$

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

I = $\left[-4cos\frac{x}{4}\right]_0^{2\pi}+\left[4sin\frac{x}{4}\right]_0^{2\pi}$

I = [-4cosπ/2 + 4cos0] + [4sinπ/2 – 4sin0]

I = 0 + 4 + 4 – 0

I = 8

Therefore, the value of $\int_0^{2\pi}\sqrt{1+sin\frac{x}{2}}dx$ is 8.

问题 63。 $\int_0^{\frac{\pi}{4}}(tanx+cotx)^{-2}dx$

解决方案：

We have,

I = $\int_0^{\frac{\pi}{4}}(tanx+cotx)^{-2}dx$

I = $\int_0^{\frac{\pi}{4}}\frac{1}{(tanx+cotx)^2}dx$

I = $\int_0^{\frac{\pi}{4}}\frac{1}{(\frac{sinx}{cosx}+\frac{cosx}{sinx})^2}dx$

I = $\int_0^{\frac{\pi}{4}}\frac{1}{(\frac{sin^2x+cos^2x}{sinxcosx})^2}dx$

I = $\int_0^{\frac{\pi}{4}}\frac{1}{(\frac{1}{sinxcosx})^2}dx$

I = $\int_0^{\frac{\pi}{4}}sin^2xcos^2xdx$

I = $\int_0^{\frac{\pi}{4}}sin^2x(1-sin^2x)dx$

I = $\int_0^{\frac{\pi}{4}}(sin^2x-sin^4x)dx$

I = $\int_0^{\frac{\pi}{4}}sin^2xdx-\int_0^{\frac{\pi}{4}}sin^4xdx$

I = $\left[\frac{x}{2}-\frac{cosxsinx}{2}\right]^{\frac{\pi}{4}}_0-\left[\frac{3}{4}(\frac{x}{2}-\frac{cosxsinx}{2})-\frac{cosxsin^3x}{4}\right]^{\frac{\pi}{4}}_0$

I = $(\frac{\pi}{8}-\frac{cos\frac{\pi}{4}sin\frac{\pi}{4}}{2})-\left(\frac{3}{4}(\frac{\pi}{8}-\frac{cos\frac{\pi}{4}sin\frac{\pi}{4}}{2})-\frac{cos\frac{\pi}{4}sin^3\frac{\pi}{4}}{4}\right)$

I = (π/8 – 1/4) – (3/4(π/8 – 1/4) – 1/16)

I = π/8 – 1/4 – (3π/32 – 3/16 – 1/16)

I =  π/8 – 1/4 – (3π/32 – 1/4)

I = π/8 – 1/4 – 3π/32 + 1/4

I = π/8 – 3π/32

I = (4π – 3π)/32

I = π/32 

Therefore, the value of $\int_0^{\frac{\pi}{4}}(tanx+cotx)^{-2}dx$ is π/32.

问题 64。 $\int_0^{1}xlog(1+2x)dx$

解决方案：

We have,

I = $\int_0^{1}xlog(1+2x)dx$

By using integration by parts we get,

I = $\frac{x^2log(1+2x)}{2}-\int(\frac{2}{2x+1}\int xdx)dx$

I = $\frac{x^2log(1+2x)}{2}-\int \frac{2x^2}{2(2x+1)}dx$

I = $\frac{x^2log(1+2x)}{2}-\int \frac{x^2}{2x+1}dx$

I = $\frac{x^2log(1+2x)}{2}-\int_0^1 \frac{x}{2}-\frac{1}{4}+\frac{1}{4(2x+1)}dx$

I = $\frac{x^2log(1+2x)}{2}-(\frac{x^2}{4}-\frac{x}{4}+\frac{1}{8}log|2x+1|)$

So we get,

I = $\left[\frac{x^2log(1+2x)}{2}\right]^1_0-\left[\frac{x^2}{4}-\frac{x}{4}+\frac{1}{8}log|2x+1|)\right]^1_0$

I = log3/2 – 1/8log3

I = 3/8log3

Therefore, the value of $\int_0^{1}xlog(1+2x)dx$ is 3/8log3.

问题 65。 $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tanx+cotx)^{2}dx$

解决方案：

We have,

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tanx+cotx)^{2}dx$

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tan^2x+cot^2x+2tanxcotx)dx$

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(sec^2x-1+cosec^2x-1+2)dx$

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(sec^2x+cosec^2x-2+2)dx$

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(sec^2x+cosec^2x)dx$

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}sec^2xdx+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}cosec^2xdx$

I = $\left[tanx\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}+\left[-cotx\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

I = [tanπ/3 – tanπ/6] + [-cotπ/3 + cotπ/6]

I = [√3 – 1/√3] + [- 1/√3 – √3]

I = 2[√3 – 1/√3]

I = 4/√3

Therefore, the value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tanx+cotx)^{2}dx$ is 4/√3.

问题 66。 $\int_{0}^{\frac{\pi}{4}}(a^2cos^2x+b^2sin^2x)dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}}(a^2cos^2x+b^2sin^2x)dx$

I = $\int_{0}^{\frac{\pi}{4}}(a^2(1-sin^2x)+b^2sin^2x)dx$

I = $\int_{0}^{\frac{\pi}{4}}(a^2-a^2sin^2x+b^2sin^2x)dx$

I = $\int_{0}^{\frac{\pi}{4}}[a^2-(b^2-a^2)sin^2x]dx$

I = $\int_{0}^{\frac{\pi}{4}}a^2dx-\int_{0}^{\frac{\pi}{4}}[(b^2-a^2)sin^2x]dx$

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

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

I = $\left[a^2x\right]_{0}^{\frac{\pi}{4}}+(b^2-a^2)[\frac{x}{2}+\frac{sin2x}{4}]_{0}^{\frac{\pi}{4}}$

I = $\frac{a^2\pi}{4}+(b^2-a^2)(\frac{\pi}{8}+\frac{1}{4})$

I = $\frac{a^2\pi}{4}-\frac{(b^2-a^2)\pi}{8}+\frac{b^2-a^2}{4}$

I = $\frac{(b^2-a^2)\pi}{8}+\frac{b^2-a^2}{4}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}(a^2cos^2x+b^2sin^2x)dx$ is $\frac{(b^2-a^2)\pi}{8}+\frac{b^2-a^2}{4}$.

问题 67。 $\int_{0}^{1}\frac{dx}{1+2x+2x^2+2x^3+x^4}$

解决方案：

We have,

I = $\int_{0}^{1}\frac{dx}{1+2x+2x^2+2x^3+x^4}$

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

I = $\int_{0}^{1}[\frac{-x}{2(x^2+1)}+\frac{1}{2(x+1)}+\frac{1}{2(x+1)^2}]dx$

I = $-\left[\frac{log(x^2+1)}{4}\right]^1_0+\left[\frac{log(x+1)}{2}\right]^1_0-\left[\frac{1}{2(x+1)}\right]^1_0$

I = -log2/4 + log2/2 – 1/4 + 1/2

I = log2/4 + 1/4

Therefore, the value of $\int_{0}^{1}\frac{dx}{1+2x+2x^2+2x^3+x^4}$ is log2/4 + 1/4.