第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.2 |设置 3

计算以下定积分：

问题 42。 $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$

解决方案：

We have,

I = $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$

Let 5 – 4 cos θ = t. So, we have

=> 4 sin θ dθ = dt

=> sin θ dθ = dt/4

Now, the lower limit is, θ = 0

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos 0

=> t = 5 – 4

=> t = 1

Also, the upper limit is, θ = π

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos π

=> t = 5 + 4

=> t = 9

So, the equation becomes,

I = $\int_{1}^{9}\frac{5t^{\frac{1}{4}}}{4}dt$

I = $\frac{5}{4}\int_{1}^{9}t^{\frac{1}{4}}dt$

I = $\frac{5}{4}\left[\frac{t^{\frac{5}{4}}}{\frac{5}{4}}\right]_{1}^{9}$

I = $\frac{5}{4}\left[\frac{4}{5}t^{\frac{5}{4}}\right]_{1}^{9}$

I = $\left[t^{\frac{5}{4}}\right]_{1}^{9}$

I = $9^{\frac{5}{4}}-1^{\frac{5}{4}}$

I = $3^{\frac{5}{2}}-1$

I = 9√3 – 1

Therefore, the value of $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$ is 9√3 – 1.  

问题 43。 $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{\cos^{3}2\theta} d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{(\cos2\theta)(\cos^22\theta)} d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\frac{\tan2\theta}{\cos^22\theta} d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\tan2\theta \sec^22\theta d\theta$

Let tan 2θ = t. So, we have

=> 2 sec² 2θ dθ = dt

=> sec² 2θ dθ = dt/2

Now, the lower limit is, θ = 0

=> t = tan 2θ

=> t = tan 0

=> t = 0

Also, the upper limit is, θ = π/6

=> t = tan 2θ

=> t = tan π/3

=> t = √3

So, the equation becomes,

I = $\frac{1}{2}\int_{0}^{\sqrt{3}}tdt$

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

I = $\frac{1}{2}\left[\frac{3}{2}-0\right]$

I = $\frac{3}{4}$

Therefore, the value of $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$ is $\frac{3}{4}$.

问题 44。 $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$

解决方案：

We have,

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

Let $x^{\frac{2}{3}}$ = t. So, we have

=> $\frac{3}{2}\sqrt{x}dx$ = dt

Now, the lower limit is, x = 0

=> t = $x^{\frac{2}{3}}$

=> t = $0^{\frac{2}{3}}$

=> t = 0

Also, the upper limit is, x = $\pi^{\frac{3}{2}}$

=> t = $x^{\frac{2}{3}}$

=> t = $(\pi^{\frac{3}{2}})^{\frac{2}{3}}$

=> t = π

So, the equation becomes,

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

I = $\frac{1}{3}\int_{0}^{\pi}(1+cos2t)dt$

I = $\frac{1}{3}\left[t+\frac{sin2t}{t}\right]_{0}^{\pi}$

I = $\frac{1}{3}\left[\pi+0-0-0\right]$

I = $\frac{\pi}{3}$

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

问题 45。 $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

解决方案：

We have,

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

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

=> 1/x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + log x

=> t = 1 + log 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I = $\int_{1}^{1+\log2}\frac{1}{t^2}dt$

I = $\left[\frac{-1}{t}\right]_{1}^{1+\log2}$

I = $\frac{-1}{1+\log2}+1$

I = $\frac{\log2}{1+\log2}$

Therefore, the value of $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$ is $\frac{\log2}{1+\log2}$.

问题 46。 $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$

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

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}(1-t^2)^2dt$

I = $\int_{0}^{1}(1+t^4-2t^2)dt$

I = $\left[t-\frac{2}{3}t^3+\frac{t^5}{5}\right]_{0}^{1}$

I = $1-\frac{2}{3}+\frac{1}{5}$

I = $\frac{8}{15}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$ is $\frac{8}{15}$.

问题 47。 $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$

解决方案：

We have,

I = $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$

Let 30 – x^3/2 = t. So, we have

=> $-\frac{3}{2}\sqrt{x}dx$ = dt

=> $\frac{3}{2}\sqrt{x}dx$ = – dt

Now, the lower limit is, x = 4

=> t = 30 – x^3/2

=> t = 30 – 4^3/2

=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x^3/2

=> t = 30 – 9^3/2

=> t = 30 – 27

=> t = 3

So, the equation becomes,

I = $\int_{22}^{3}\frac{-2}{3t^2}dt$

I = $\frac{2}{3}\int_{3}^{22}\frac{1}{t^2}dt$

I = $\frac{2}{3}\left[\frac{-1}{t}\right]_{3}^{22}$

I = $\frac{2}{3}\left[\frac{-1}{22}+\frac{1}{3}\right]$

I = $\frac{2}{3}(\frac{19}{66})$

I = $\frac{19}{99}$

Therefore, the value of $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$ is $\frac{19}{99}$.

问题 48。 $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$

解决方案：

We have,

I = $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$

Let cos x = t. So, we have

=> – sin x dx = dt

=> 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

=> t = cos x

=> t = cos π

=> t = –1

So, the equation becomes,

I = $\int_{0}^{\pi}sin^2x(1+2cosx)(1+cosx)^2.sinxdx$

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

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

I = $\int_{-1}^{1}(1+4t+4t^2-2t^3-5t^4-2t^5)dt$

I = $\left[t+2t^2+\frac{4}{3}t^3-\frac{1}{2}t^4-t^5-\frac{1}{3}t^6\right]_{-1}^{1}$

I = $2+0+\frac{8}{3}-0-2-0$

I = $\frac{8}{3}$

Therefore, the value of $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$ is $\frac{8}{3}$.

问题 49。 $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$

解决方案：

We have,

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

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $2\int_{0}^{1}ttan^{-1}tdt$

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

I = $2(\frac{\pi}{4}-\frac{1}{2})$

I = $\frac{\pi}{2}-1$

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

问题 50。 $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

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

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $2\int_{0}^{1}ttan^{-1}tdt$

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

I = $2(\frac{\pi}{4}-\frac{1}{2})$

I = $\frac{\pi}{2}-1$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$ is $\frac{\pi}{2}-1$.

问题 51。 $\int_{0}^{1}(cos^{-1}x)^2dx$

解决方案：

We have,

I = $\int_{0}^{1}(cos^{-1}x)^2dx$

On using integration by parts, we get,

I = $(cos^{-1}x)^2\int_{0}^{1}dx-\int_0^1(\int dx)\frac{d}{dx}(cos^{-1}x)^2dx$

I = $\left[x(cos^{-1}x)^2\right]_{0}^{1}+2\int_0^1\frac{xcos^{-1}x}{\sqrt{1-x^2}}dx$

Let cos^-1 x = t. So, we have

=> $\frac{-1}{\sqrt{1-x^2}}dx$ = dt

Now, the lower limit is, x = 0

=> t = cos^-1 x

=> t = cos^-1 0

=> t = π/2

Also, the upper limit is, x = 1

=> t = cos^-1 x

=> t = cos^-1 1

=> t = 0

So, the equation becomes,

I = $\left[x(cos^{-1}x)^2\right]_{0}^{1}+2\int_\frac{\pi}{2}^0-tcostdt$

I = $0-0+2\int^\frac{\pi}{2}_0tcostdt$

I = $2\int^\frac{\pi}{2}_0tcostdt$

I = $t\int_{0}^{\frac{\pi}{2}}costdt-\int_0^\frac{\pi}{2}(\int costdt)\frac{dt}{dt}dt$

I = $2\left[tsint-\int sintdt\right]^\frac{\pi}{2}_0$

I = $2\left[tsint+cost\right]^\frac{\pi}{2}_0$

I = $2(\frac{\pi}{2}-1)$

I = π – 2

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

问题 52。 $\int_{0}^{a}sin^{-1}\sqrt{\frac{x}{a+x}}dx$

解决方案：

We have,

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

Let x = a tan² t. So, we have

=> dx = 2a tan t sec² t dt

Now, the lower limit is, x = 0

=> a tan² t = x

=> a tan² t = 0

=> tan t = 0

=> t = 0

Also, the upper limit is, x = a

=> a tan² t = x

=> a tan² t = a

=> tan² t = 1

=> tan t = 1

=> t = π/4

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{atan^2t}{a+atan^2t}}(2atantsec^2t)dt$

I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{atan^2t}{a(1+tan^2t)}}(2atantsec^2t)dt$

I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{tan^2t}{sec^2t}}(2atantsec^2t)dt$

I = $2a\int_{0}^{\frac{\pi}{4}}sin^{-1}(sint)(tantsec^2t)dt$

I = $2a\int_{0}^{\frac{\pi}{4}}t(tantsec^2t)dt$

I = $2a\left[\frac{ttan^2t}{2}-\int \frac{tan^2t}{2}dt\right]_0^\frac{\pi}{4}$

I = $\left[attan^2t-\frac{2a}{2}\int (sec^2t-1)dt\right]_0^\frac{\pi}{4}$

I = $\left[attan^2t-atant+at\right]_0^\frac{\pi}{4}$

I = $\frac{a\pi}{4}tan^2\frac{\pi}{4}-atan\frac{\pi}{4}+a\frac{\pi}{4}$

I = $\frac{a\pi}{4}-a+\frac{a\pi}{4}$

I = $\frac{a\pi}{2}-a$

I = $a(\frac{\pi}{2}-1)$

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

问题 53。 $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$

解决方案：

We have,

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

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

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

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

I = $\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}cot\frac{x}{2}\cosec^2\frac{x}{2}dx$

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

=> $\frac{-1}{2}cosec^2\frac{x}{2}dx$ = dt

Now, the lower limit is, x = π/3

=> t = cot x/2

=> t = cot π/6

=> t = √3

Also, the upper limit is, x = π/2

=> t = cot x/2

=> t = cot π/4

=> t = 1

So, the equation becomes,

I = $-\int_{\sqrt{3}}^{1}tdt$

I = $\int^{\sqrt{3}}_{1}tdt$

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

I = $\frac{3}{2}-\frac{1}{2}$

I = 1

Therefore, the value of $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$ is 1.

问题 54。 $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$

解决方案：

We have,

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

Let x² = a² cos 2t. So, we have

=> 2x dx = – 2a² sin 2t dt

Now, the lower limit is, x = 0

=> a² cos 2t = x²

=> a² cos 2t = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = a

=> a² cos 2t = x²

=> a² cos 2t = a²

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

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

I = $-a^2\int_{\frac{\pi}{4}}^{0}\frac{sint}{cost}sin2tdt$

I = $a^2\int^{\frac{\pi}{4}}_{0}\frac{sint}{cost}sin2tdt$

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

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

I = $a^2\left[t-\frac{sin2t}{2}\right]^{\frac{\pi}{4}}_{0}$

I = $a^2(\frac{\pi}{4}-\frac{1}{2})$

Therefore, the value of $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$ is $a^2(\frac{\pi}{4}-\frac{1}{2})$.

问题 55。 $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$

解决方案：

We have,

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

Let x = a cos 2t. So, we have

=> dx = –2a sin 2t

Now, the lower limit is, x = –a

=> a cos 2t = x

=> a cos 2t = –a

=> cos 2t = –1

=> 2t = π

=> t = π/2

Also, the upper limit is, x = a

=> a cos 2t = x

=> a cos 2t = a

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = $\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{a-acos2t}{a+acos2t}}(-2asin2t)dt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{a(1-cos2t)}{a(1+cos2t)}}sin2tdt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{1-cos2t}{1+cos2t}}sin2tdt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{2sin^2t}{2cos^2t}}sin2tdt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\frac{sint}{cost}(2sintcost)dt$

I = $-4a\int_{\frac{\pi}{2}}^{0}sin^2tdt$

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

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

I = $2a\left[t-\frac{sin2t}{2}\right]^{\frac{\pi}{2}}_{0}$

I = $2a\left[\frac{\pi}{2}-0-0+0\right]$

I = πa

Therefore, the value of $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$ is πa.

问题 56。 $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$

解决方案：

We have,

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

Let cos x = t. So, we have

=> – sin x dx = dt

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

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

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

I = $\left[-log|1+t|+2log|t+2|\right]_{0}^{1}$

I = – log 2 + 2 log 3 + 0 – 2 log 2

I = 2 log 3 – 3 log 2

I = log 9 – log 8

I = log 9/8

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$ is log 9/8.

问题 57。 $\int_{0}^{\frac{\pi}{2}}\frac{tanx}{1+m^2tan^2x}dx$

解决方案：

We have,

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

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

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

Let sin² x = t. So, we have

=> 2 sin x cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin² x

=> t = sin² 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin² x

=> t = sin² π/2

=> t = 1

So, the equation becomes,

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

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

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

I = $\frac{1}{2(m^2-1)}\left[\log|m^2-1+1|-log1\right]$

I = $\frac{1}{2(m^2-1)}\left[\log m^2-log1\right]$

I = $\frac{\log m^2}{2(m^2-1)}$

I = $\frac{2\log m}{2(m^2-1)}$

I = $\frac{\log m}{m^2-1}$

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

问题 58。 $\int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)(\sqrt{1-x^2})}dx$

解决方案：

We have,

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

Let x = sin t. So, we have

=> dx = cos t dt

Now, the lower limit is, x = 0

=> sin t = x

=> sin t = 0

=> t = 0

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

=> sin t = x

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)(\sqrt{1-sin^2t})}(cost)dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)(\sqrt{cos^2t})}(cost)dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)cost}(cost)dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{1+sin^2t}dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{sec^2t}{1+2tan^2t}dt$

Let tan t = u. So, we have

=> sec² t dt = du

Now, the lower limit is, t = 0

=> u = tan t

=> u = tan 0

=> t = 0

Also, the upper limit is, t = π/6

=> u = tan t

=> u = tan π/6

=> t = 1/√3

So, the equation becomes,

I = $\int_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+2u^2}du$

I = $\frac{1}{\sqrt{2}}\left[tan^{-1}(\sqrt{2}u)\right]_{0}^{\frac{1}{\sqrt{3}}}$

I = $\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}$

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

问题 59。 $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$

解决方案：

We have,

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

Let 1/x² – 1 = t. So, we have

=> –2/x³ dx = dt

Now, the lower limit is, x = 1/3

=> t = 1/x² – 1

=> t = 9 – 1 

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x² – 1

=> t = 1 – 1

=> t = 0

So, the equation becomes,

I = $\frac{-1}{2}\int_{8}^{0}t^{\frac{1}{3}}dt$

I = $\frac{1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]_{0}^{8}$

I = $\frac{3}{8}\left[t^{\frac{4}{3}}\right]_{0}^{8}$

I = $\frac{3}{8}(8^{\frac{4}{3}})$

I = $\frac{3}{8}(16)$

I = 6

Therefore, the value of $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$ is 6.

问题 60。 $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$

I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^2xsec^2x}{tan^6x+2tan^3x+1}dx$

Let tan x = t. So, we have

=> sec² x dx = dt 

Now, the lower limit is, x = 0

=> t = tan t

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan t

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{t^2}{t^6+2t^3+1}dt$

Let t³ = u. So, we have

=> 3t² dt = du

=> t² dt = du/3

Now, the lower limit is, t = 0

=> u = t³

=> u = 0³

=> u = 0

Also, the upper limit is, t = 1

=> u = t³

=> u = 1³

=> u = 1

So, the equation becomes,

I = $\frac{1}{3}\int_{0}^{1}\frac{1}{u^2+2u+1}du$

I = $\frac{1}{3}\int_{0}^{1}\frac{1}{(u+1)^2}du$

I = $\frac{1}{3}\left[\frac{-1}{u+1}\right]_{0}^{1}$

I = $\frac{1}{3}\left[\frac{-1}{1+1}+\frac{1}{1+0}\right]$

I = $\frac{1}{3}(\frac{-1}{2}+1)$

I = $\frac{1}{3}(\frac{1}{2})$

I = $\frac{1}{6}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$ is $\frac{1}{6}$.

问题 61。 $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}tan^2xcos^2xdx$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}sin^2xdx$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx(1-cos^2x)}sin^2xdx$

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

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 = $-\int_{1}^{0}\sqrt{t}(1-t^2)dt$

I = $\int_{0}^{1}\sqrt{t}(1-t^2)dt$

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

I = $\left[\frac{2t^{\frac{3}{2}}}{3}-\frac{2t^{\frac{7}{2}}}{7}\right]_{0}^{1}$

I = $\frac{2}{3}-\frac{2}{7}$

I = $\frac{8}{21}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$ is $\frac{8}{21}$.

问题 62。 $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$

解决方案：

We have,

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

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

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

Let cos x/2 + sin x/2 = t. So, we have

=> (cos x/2 – sin x/2) dx = 2 dt

Now, the lower limit is, x = 0

=> t = cos x/2 + sin x/2

=> t = cos 0 + sin 0

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x/2 + sin x/2

=> t = cos π/2 + sin π/2

=> t = 1/√2 + 1/√2

=> t = √2

So, the equation becomes,

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

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

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

I = $\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right]$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$ is $\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right]$.