第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.2 |设置 2

计算以下定积分：

问题 21。 $\int_{0}^{\pi}\frac{sinx}{sinx+cosx}dx$

解决方案：

We have,

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

Let sin x = A (sin x + cos x) + B$\frac{d}{dx}(sinx+cosx)$

=> sin x = A (sin x + cos x) + B (cos x – sin x)

=> sin x = sin x (A – B) + cos x (A + B)

On comparing both sides, we get

A – B = 1 and A + B = 0

On solving, we get A = 1/2 and B = –1/2.

Therefore, the expression becomes,

I = $\frac{1}{2}\int_0^\pi dx-\frac{1}{2}\int_{0}^{\pi}\frac{cosx-sinx}{sinx+cosx}dx$

I = $\left[\frac{x}{2}\right]_0^\pi-\frac{1}{2}\left[\log(sinx+cosx)\right]_{0}^{\pi}$

I = $\frac{\pi}{2}-\frac{1}{2}(0)$

I = $\frac{\pi}{2}$

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

问题 22。 $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$

解决方案：

We have,

I = $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$

On putting cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$ and sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$, we get,

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

I = $\int_{0}^{\pi}\frac{1+tan^2\frac{x}{2}}{3(1+tan^2\frac{x}{2})+4tan\frac{x}{2}+1-tan^2\frac{x}{2}}dx$

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

I = $\int_{0}^{\pi}\frac{sec^2\frac{x}{2}}{2tan^2\frac{x}{2}+4tan\frac{x}{2}+4}dx$

Let tan x/2 = t. So, we have

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I = $\int_{0}^{\infty}\frac{2}{2t^2+4t+4}dt$

I = $\int_{0}^{\infty}\frac{1}{t^2+2t+2}dt$

I = $\int_{0}^{\infty}\frac{1}{(t+1)^2+1}dt$

I = $\left[tan^{-1}(t+1)\right]^\infty_0$

I = $tan^{-1}\infty-tan^{-1}1$

I = $\frac{\pi}{2}-\frac{\pi}{4}$

I = $\frac{\pi}{4}$

Therefore, the value of $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$ is $\frac{\pi}{4}$.

问题 23。 $\int_{0}^{1}tan^{-1}xdx$

解决方案：

We have,

I = $\int_{0}^{1}tan^{-1}xdx$

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

I = $\left[xtan^{-1}x\right]_{0}^{1}-\int_{0}^1\frac{x}{1+x^2}dx$

I = $\left[xtan^{-1}x\right]_{0}^{1}-\frac{1}{2}\int_{0}^1\frac{2x}{1+x^2}dx$

I = $\left[xtan^{-1}x\right]_{0}^{1}-\frac{1}{2}\left[log(1+x^2)\right]_{0}^1$

I = $(\frac{\pi}{4}-0)-\frac{1}{2}\left[log(1+1)-log(1+0)\right]$

I = $\frac{\pi}{4}-\frac{1}{2}\left[log2-log1\right]$

I = $\frac{\pi}{4}-\frac{1}{2}\left[log2-0\right]$

I = $\frac{\pi}{4}-\frac{\log2}{2}$

Therefore, the value of $\int_{0}^{1}tan^{-1}xdx$ is $\frac{\pi}{4}-\frac{\log2}{2}$.

问题 24。 $\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx$

解决方案：

We have,

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

Let sin^–1 x = t. So, we have

=> $\frac{1}{\sqrt{1+x^2}}dx$ = dt

Now, the lower limit is, x = 0

=> t = sin^–1 x

=> t = sin^–1 0

=> t = 0

Also, the upper limit is, x = 1/2

=> t = sin^–1 x

=> t = sin^–1 1/2

=> t = π/6

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{6}}tsintdt$

I = $t\int_{0}^{\frac{\pi}{6}}sintdt-\int_{0}^{\frac{\pi}{6}}(\int sintdt) \frac{d}{dt}(t)dt$

I = $\left[-tcost\right]_{0}^{\frac{\pi}{6}}+\int_{0}^{\frac{\pi}{6}}costdt$

I = $\left[-tcost\right]_{0}^{\frac{\pi}{6}}+\left[sint\right]_{0}^{\frac{\pi}{6}}$

I = $\left[-\frac{\pi}{6}(\frac{\sqrt{3}}{2})+0\right]+\left[sin\frac{\pi}{6}-0\right]$

I = $\frac{-\sqrt{3}\pi}{12}+\frac{1}{2}$

I = $\frac{1}{2}-\frac{\sqrt{3}\pi}{12}$

Therefore, the value of $\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx$ is $\frac{1}{2}-\frac{\sqrt{3}\pi}{12}$.

问题 25。 $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$

I = $\int_{0}^{\frac{\pi}{4}}(\sqrt{\frac{sinx}{cosx}}+\sqrt{\frac{cosx}{sinx}})dx$

I = $\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{sinxcosx}})dx$

I = $\sqrt{2}\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{2sinxcosx}})dx$

I = $\sqrt{2}\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{1-(sinx-cosx)^2}})dx$

Let sinx – cosx = t. So, we have

=> (cos x + sin x) dx = dt

Now, the lower limit is, x = 0

=> t = sinx – cosx

=> t = sin 0 – cos 0

=> t = 0 – 1

=> t = –1

Also, the upper limit is, x = π/4

=> t = sinx – cosx

=> t = sin π/4 – cos π/4

=> t = sin π/4 – sin π/4

=> t = 0

So, the equation becomes,

I = $\sqrt{2}\int_{-1}^{0}\frac{1}{\sqrt{1-t^2}}dt$

I = $\sqrt{2}\left[sin^{-1}t\right]_{-1}^{0}$

I = $\sqrt{2}\left[sin^{-1}0-sin^{-1}(-1)\right]$

I = $\sqrt{2}\left[0+sin^{-1}(1)\right]$

I = $\sqrt{2}(\frac{\pi}{2})$

I = $\frac{\pi}{\sqrt{2}}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$ is $\frac{\pi}{\sqrt{2}}$.

问题 26。 $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$

I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{2cos^2x}dx$

I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^3xsec^2x}{2}dx$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{4}}tan^3xsec^2xdx$

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan x

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = $\frac{1}{2}\int_{0}^{1}t^3dt$

I = $\frac{1}{2}\left[\frac{t^4}{4}\right]_{0}^{1}$

I = $\frac{1}{2}(\frac{1}{4}-0)$

I = $\frac{1}{8}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$ is $\frac{1}{8}$.

问题 27。 $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$

解决方案：

We have,

I = $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$

On putting cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$, we get

I = $\int_{0}^{\pi}\frac{1}{5+\frac{3(1-tan^2\frac{x}{2})}{1+tan^2\frac{x}{2}}}dx$

I = $\int_{0}^{\pi}\frac{1+tan^2\frac{x}{2}}{5(1+tan^2\frac{x}{2})+3(1-tan^2\frac{x}{2})}dx$

I = $\int_{0}^{\pi}\frac{sec^2\frac{x}{2}}{5(1+tan^2\frac{x}{2})+3(1-tan^2\frac{x}{2})}dx$

Let tan x/2 = t. So, we have

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I = $\int_{0}^{\infty}\frac{1}{5(1+t^2)+3(1-t^2)}dt$

I = $\int_{0}^{\infty}\frac{1}{5+5t^2+3-3t^2}dt$

I = $\int_{0}^{\infty}\frac{1}{8+2t^2}dt$

I = $\frac{1}{2}\int_{0}^{\infty}\frac{1}{4+t^2}dt$

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

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

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

I = $\frac{1}{2}(\frac{\pi}{2})$

I = $\frac{\pi}{4}$

Therefore, the value of $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$ is $\frac{\pi}{4}$.

问题 28。 $\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx$

解决方案：

We have,

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

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

I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2x}{a^2tan^2x+b^2}dx$

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

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

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

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

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

I = $\frac{1}{a^2}(\frac{a}{b})(\frac{\pi}{2})$

I = $\frac{\pi}{2ab}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx$ is $\frac{\pi}{2ab}$.

问题 29。 $\int_{0}^{\frac{\pi}{2}}\frac{x+sinx}{1+cosx}dx$

解决方案：

We have,

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

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

I = $\int_{0}^{\frac{\pi}{2}}\frac{xsec^2\frac{x}{2}}{2}+tan\frac{x}{2}dx$

I = $\left[xtan\frac{x}{2}-\int_{0}^{\frac{\pi}{2}}tan\frac{x}{2}dx+\int_{0}^{\frac{\pi}{2}}tan\frac{x}{2}dx\right]^{\frac{\pi}{2}}_0$

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

I = $\frac{\pi}{2}tan\frac{\pi}{4}-0$

I = $\frac{\pi}{2}$

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

问题 30。 $\int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx$

解决方案：

We have,

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

Let tan^–1 x = t. So, we have

=> $\frac{1}{1+x^2}dx$ = dt

Now, the lower limit is, x = 0

=> t = tan^–1 x

=> t = tan^–1 0

=> t = 0

Also, the upper limit is, x = 1

=> t = tan^–1 x

=> t = tan^–1 1

=> t = π/4

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{4}}tdt$

I = $\left[\frac{t^2}{2}\right]_{0}^{\frac{\pi}{4}}$

I = $\frac{1}{2}(\frac{\pi^2}{16})-0$

I = $\frac{\pi^2}{32}$

Therefore, the value of $\int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx$ is $\frac{\pi^2}{32}$.

问题 31。 $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$

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

I = $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{4-(cosx-sinx)^2}dx$

I = $\left[\frac{1}{4}log|\frac{2+sinx-cosx}{2-sinx+cosx}|\right]_{0}^{\frac{\pi}{4}}$

I = $\frac{1}{4}\left[log|\frac{2+sin\frac{\pi}{4}-cos\frac{\pi}{4}}{2-sin\frac{\pi}{4}+cos\frac{\pi}{4}}|-log\frac{2+0-1}{2-0+1}\right]$

I = $\frac{1}{4}\left(log|\frac{2}{2}|-log\frac{1}{3}\right)$

I = $\frac{1}{4}\left(log1-log\frac{1}{3}\right)$

I = $\frac{1}{4}\left(-log\frac{1}{3}\right)$

I = $-\frac{log\frac{1}{3}}{4}$

I = $\frac{log3}{4}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$ is $\frac{log3}{4}$.

问题 32。 $\int_{0}^{1}xtan^{-1}xdx$

解决方案：

We have,

I = $\int_{0}^{1}xtan^{-1}xdx$

On using integration by parts, we get

I = $tan^{-1}x\int_{0}^{1}xdx-\int_0^1(\int xdx)\frac{d}{dx}(tan^{-1}x)dx$

I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{x^2}{1+x^2}dx$

I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2-1}{1+x^2}dx$

I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx$

I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx$

I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx$

I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\left[x\right]_0^1+\frac{1}{2}\left[tan^{-1}x\right]_0^1$

I = $(\frac{\pi}{8}-0)-\frac{1}{2}(1-0)+\frac{1}{2}\left[tan^{-1}1-tan^{-1}0\right]$

I = $\frac{\pi}{8}-\frac{1}{2}+\frac{1}{2}(\frac{\pi}{4}-0)$

I = $\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$

I = $\frac{\pi}{4}-\frac{1}{2}$

Therefore, the value of $\int_{0}^{1}xtan^{-1}xdx$ is $\frac{\pi}{4}-\frac{1}{2}$.

问题 33。 $\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx$

解决方案：

We have,

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

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

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

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

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

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

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

I = $-\frac{1}{2}\log|\frac{1^2-1+1}{1^2+1+1}|+\frac{1}{2}\log|\frac{0^2-0+1}{0^2+0+1}|$

I = $-\frac{1}{2}\log\frac{1}{3}+\frac{1}{2}\log1$

I = $-\frac{1}{2}\log\frac{1}{3}+\frac{1}{2}(0)$

I = $-\frac{1}{2}\log\frac{1}{3}$

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

I = $\log3^{\frac{1}{2}}$

I = $\frac{1}{2}\log3$

Therefore, the value of $\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx$ is $\frac{1}{2}\log3$.

问题 34。 $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$

解决方案：

We have,

I = $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$

Let 1 + x² = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + x²

=> t = 1 + 0²

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π

=> t = 1 + x²

=> t = 1 + 1²

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = $\int_{0}^{2}\frac{12(2x)(x^2)}{(1+x^2)^4}dx$

I = $\int_{1}^{2}\frac{12(t-1)}{t^4}dt$

I = $12\int_{1}^{2}(\frac{t}{t^4}-\frac{1}{t^4})dt$

I = $12\int_{1}^{2}(\frac{1}{t^3}-\frac{1}{t^4})dt$

I = $12\left[\frac{-1}{2t^2}-\frac{1}{3t^3}\right]_{1}^{2}$

I = $12\left[\frac{-1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\right]$

I = $12\left[\frac{-3+1+12-8}{24}\right]$

I = $12(\frac{2}{24})$

I = 1

Therefore, the value of $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$ is 1.

问题 35。 $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$

解决方案：

We have,

I = $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$

Let x – 4 = t³. So, we have

=> dx = 3t² dt

Now, the lower limit is, x = 4

=> t³ = x – 4

=> t³ = 4 – 4

=> t³ = 0

=> t = 0

Also, the upper limit is, x = 12

=> t³ = x – 4

=> t³ = 12 – 4

=> t³ = 8

=> t = 2

So, the equation becomes,

I = $\int_{0}^{2}(t^3+4)t(3t^2)dt$

I = $\int_{0}^{2}3t^3(t^3+4)dt$

I = $3\int_{0}^{2}(t^6+4t^3)dt$

I = $3\left[\frac{t^7}{7}+t^4\right]_{0}^{2}$

I = $3\left[\frac{128}{7}+16-0-0\right]$

I = $\frac{720}{7}$

Therefore, the value of $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$ is $\frac{720}{7}$.

问题 36。 $\int_{0}^{\frac{\pi}{2}}x^2sinxdx$

解决方案：

We have,

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

On using integration by parts, we get

I = $x^2\int sinxdx-\int(\int sinxdx)\frac{d}{dx}(x^2)dx$

I = $x^2cosx-\int 2xcosxdx$

I = $x^2cosx+2[x\int cosxdx-\int(\int cosxdx)\frac{dx}{dx}dx]$

I = $x^2cosx+2[xsinxdx-\int sinxdx]$

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

I = π + 0 – 0 – 0 – 2

I = π – 2

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}x^2sinxdx$ is π – 2.

问题 37。 $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx$

解决方案：

We have,

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

Let x = cos 2t. So, we have

=> dx = – 2 sin 2t dt

Now, the lower limit is, x = 0

=> cos 2t = x

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = 1

=> cos 2t = x

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

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

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

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

I = $4\int^{\frac{\pi}{4}}_{0}sin^2tdt$

I = $2\int^{\frac{\pi}{4}}_{0}2sin^2tdt$

I = $2\int^{\frac{\pi}{4}}_{0}(1-cos2t)dt$

I = $2\left[t-\frac{sin^2t}{2}\right]^{\frac{\pi}{4}}_{0}$

I = $2\left[\frac{\pi}{4}-\frac{1}{2}\right]$

I = $\frac{\pi}{2}-1$

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

问题 38。 $\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}dx$

解决方案：

We have,

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

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

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

Let x + 1/x = t. So, we have

=> (1 – 1/x²)dx = dt

Now, the lower limit is, x = 0

=> t = x + 1/x

=> t = ∞

Also, the upper limit is, x = 1

=> t = x + 1/x

=> t = 1 + 1

=> t = 2

So, the equation becomes,

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

I = $\int^{\infty}_{2}\frac{dt}{t^2}$

I = $\left[\frac{-1}{t}\right]^{\infty}_{2}$

I = $\frac{-1}{\infty}+\frac{1}{2}$

I = $\frac{1}{2}$

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

问题 39。 $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$

解决方案：

We have,

I = $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$

Let x⁵ + 1 = t. So, we have

=> 5x⁴ dx = dt

Now, the lower limit is, x = –1

=> t = x⁵ + 1

=> t = (–1)⁵ + 1

=> t = –1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x⁵ + 1

=> t = (1)⁵ + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = $\int_{0}^{2}\sqrt{t}dt$

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

I = $\frac{2}{3}(2^{\frac{3}{2}})$

I = $\frac{2}{3}(2\sqrt{2})$

I = $\frac{4\sqrt{2}}{3}$

Therefore, the value of $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$ is $\frac{4\sqrt{2}}{3}$.

问题 40。 $\int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx$

解决方案：

We have,

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

I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2x}{sec^2x(sec^2x+3tan^2x)}dx$

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

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

I = $\frac{-1}{3}\int_{0}^{\infty}(\frac{1}{1+t^2}-\frac{1}{1+4t^2})dt$

I = $\frac{-1}{3}\left[tan^{-1}t-2tan^{-1}2t\right]_{0}^{\infty}$

I = $\frac{-1}{3}(\frac{\pi}{2})$

I = $\frac{-\pi}{6}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx$ is $\frac{-\pi}{6}$.

问题 41。 $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$

Let sin 2t = u. So, we have

=> 2 cos 2t dt = du

=> cos 2t dt = du/2

Now, the lower limit is, x = 0

=> u = sin 2t

=> u = sin 0

=> u = 0

Also, the upper limit is, x = π/4

=> u = sin 2t

=> u = sin π/2

=> u = 1

So, the equation becomes,

I = $\frac{1}{2}\int_{0}^{1}u^3du$

I = $\frac{1}{2}\left[\frac{u^4}{4}\right]_{0}^{1}$

I = $\frac{1}{2}(\frac{1}{4})$

I = $\frac{1}{8}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$ is $\frac{1}{8}$.