第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.1 |设置 2

计算以下定积分：

问题 23。 $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx$

解决方案：

We have,

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

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

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

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

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

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

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

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

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

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

I = [(a² + b²)/2][π/2]

I = π(a² + b²)/4

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx$is π(a² + b²)/4.

问题 24。 $\int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx$

解决方案：

We have,

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

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

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

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

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

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

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

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

I = 2[sinπ/4 – cosπ/4 – 0 + 1]

I = 2[1/√2 – 1/√2 – 0 + 1]

I = 2 (1)

I = 2

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

问题 25。 $\int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx$

解决方案：

We have,

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

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

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

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

I = $\sqrt{2}\left[2sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}$

I = $2\sqrt{2}\left[sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}$

I = 2√2[sinπ/4 – sin0]

I = 2√2[1/√2- sin0]

I = 2√2[1/√2] 

I = 2

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

问题 26。 $\int_{0}^{\frac{\pi}{2}} xsinxdx$

解决方案：

We have,

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

By using integration by parts, we get,

I = x ∫sinxdx – ∫(∫sin x (1)dx)dx

I = -xcosx – ∫(∫sin xdx)dx

I = -xcosx + ∫cosxdx

I = -xcosx + sinx

So we get,

I = $\left[-xcosx+sinx\right]^{\frac{\pi}{2}}_0$

I = [-π/2cosπ/2 + sinπ/2 + 0 – 0]

I = 0 + 1 + 0 – 0 

I = 1

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

问题 27。 $\int_{0}^{\frac{\pi}{2}} xcosxdx$

解决方案：

We have,

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

By using integration by parts, we get,

I = x∫cosxdx – ∫(∫cos x (1)dx)dx

I = xsinx – ∫(∫cosxdx)dx

I = xsinx – ∫sinxdx

I = x sin x + cos x

So we get,

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

I = [π/2sinπ/2 + cosπ/2 – 0 – cos0]

I = π/2 + 0 – 0 – 1 

I = π/2 – 1  

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} xcosxdx$is π/2 – 1.

问题 28。 $\int_{0}^{\frac{\pi}{2}} x^2cosxdx$

解决方案：

We have,

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

By using integration by parts, we get,

I = x²sinx – ∫(2x∫(cosx)dx)dx

I = x²sinx – ∫(2xsinx)dx

I = x²sinx – 2[-xcosx – ∫(1∫sinxdx)dx]

I = x²sinx – 2[-xcosx + ∫sinxdx]

I = x²sinx – 2[-xcosx + sinx]

I = x²sinx + 2xcosx – 2sinx

So we get,

I = $\left[x^2sinx+2xcosx-2sinx\right]^{\frac{\pi}{2}}_0$

I = [(π/2)²sinπ/2 + 2(π/2)cosπ/2 – 2sinπ/2 – 0 – 0 + sin0]

I = [π²/4 + 0 – 2 – 0 – 0 + 0]

I = π²/4 – 2 

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} x^2cosxdx$ is π²/4 – 2.

问题 29。 $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$

By using integration by parts, we get,

I = -x²cosx – ∫(2x∫sinxdx)dx

I = -x²cosx + ∫(2xcosx)dx

I = -x²cosx + 2[xsinx – ∫(∫cosxdx)dx]

I = -x²cosx + 2[xsinx – ∫sinxdx]

I = -x²cosx + 2[xsinx + cosx]

I = -x²cosx + 2xsinx + 2cosx

So we get,

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

I = -(π/4)²cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 – 0 – 2

I = –π²/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2

I = –π²/16√2 + π/2√2 + √2 –  2

Therefore, the value of $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$is -π²/16√2 + π/2√2 + √2 –  2.

问题 30。 $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$

By using integration by parts, we get,

I = 1/2x²sin2x – ∫(2x∫cos2xdx)dx

I = 1/2x²sin2x – ∫(xsin2x)dx

I = 1/2x²sin2x – [-1/2xcos2x – ∫(∫sin2xdx)dx]

I = 1/2x²sin2x – [-1/2xcos2x + ∫1/2 cos2xdx]

I = 1/2x²sin2x – [-1/2xcos2x + 1/4sin2xdx]

I = 1/2x²sin2x + 1/2xcos2x – 1/4sin2xdx

So we get,

I = $\left[\frac{1}{2}x^2sin2x+\frac{1}{2}xcos2x-\frac{1}{4}sin2xdx\right]^{\frac{\pi}{2}}_0$

I = [1/2(π²/4)sinπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]

I = -π/4

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$is -π/4.

问题 31。 $\int_{0}^{\frac{\pi}{2}} x^2cos^2xdx$

解决方案：

We have,

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

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

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

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

By using integration by parts, we get,

I = 1/2[x³/3] + x²sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]

I = 1/2[x³/3] + x²sin2x/2 + xcosx/2 – sin2x/4

So we get,

I = $\left[\frac{1}{2}[\frac{x^3}{3}]+\frac{x^2sin2x}{2}+\frac{xcosx}{2}-\frac{sin2x}{4}\right]^{\frac{\pi}{2}}_0$

I = [1/6[π³/8] + 0 + 0 – π/8] 

I = π³/48 – π/8

Therefore, the value of $\int_{0}^{\frac{\pi}{2}} x^2cos^2xdx$ is π³/48 – π/8.

问题 32。 $\int_{1}^{2}logxdx$

解决方案：

We have,

I = $\int_{1}^{2}logxdx$

By using integration by parts, we get,

I = $xlogx(1)-\int x\frac{1}{x}dx$

I = xlogx – ∫1dx

I = xlogx – x

So we get,

I = $\left[xlogx-x\right]^2_1$

I = 2log2 – 2 – log1 + 1

I = 2 log 2 – 1

Therefore, the value of $\int_{1}^{2}logxdx$is 2 log 2 – 1.

问题 33。 $\int_{1}^{3}\frac{logx}{(x+1)^2}dx$

解决方案：

We have,

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

By using integration by parts, we get,

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

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

I = $-\frac{logx}{x+1}+\int (\frac{1}{x}-\frac{1}{x+1})dx$

I = $-\frac{logx}{x+1}+logx - log(x+1)$

So we get,

I = $\left[-\frac{logx}{x+1}+logx - log(x+1)\right]^3_1$

I = -log3/4 + log3 – log4 + log1/2 – log1 + log2

I = log3(1 – 1/4) – 2log2 + 0 – 0 + log2

I = 3/4log3 – log2

Therefore, the value of $\int_{1}^{3}\frac{logx}{(x+1)^2}dx$is 3/4log3 – log2.

问题 34。 $\int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx$

解决方案：

We have,

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

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

I = $\int_{1}^{e}\frac{e^x}{x}dx+\int_{1}^{e}e^xlogxdx$

By using integration by parts, we get,

I = $e^xlogx-\int_{1}^{e}e^xlogxdx+\int_{1}^{e}e^xlogxdx$

I = e^xlogx

So we get,

I = $\left[e^xlogx\right]^e_1$

I = e^eloge – e¹log1

I = e^e (1) – 0

I = e^e

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

问题 35。 $\int_{1}^{e}\frac{logx}{x}dx$

解决方案：

We have,

I = $\int_{1}^{e}\frac{logx}{x}dx$

Let log x = t, so we have,

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = e

=> t = log x

=> t = log e

=> t = 1

So, the equation becomes,

I = $\int_{1}^{e}\frac{logx}{x}dx$

I = $\int_{0}^{1}tdt$

I = $\left[\frac{t^2}{2}\right]^1_0$

I = 1/2 – 0/2

I = 1/2

Therefore, the value of $\int_{1}^{e}\frac{logx}{x}dx$is 1/2.

问题 36。 $\int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx$

解决方案：

We have,

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

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

By using integration by parts, we get,

I = $\frac{x}{logx}-\int [(\frac{-1}{(logx)^2})(\frac{1}{x})\int dx]dx -\int \frac{1}{(logx)^2}dx$

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

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

I = x/logx

So we get,

I = $\left[\frac{x}{logx}\right]^{e^2}_e$

I = $\left[\frac{e^2}{loge^2}-\frac{e}{loge}\right]$

I = $\left[\frac{e^2}{2loge}-e\right]$

I = e²/2 – e

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

问题 37。 $\int_{1}^{2}\frac{x+3}{x(x+2)}dx$

解决方案：

We have,

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

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

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

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

I = $\left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1$

I = $\left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1$

I = $\left[\frac{3}{2}logx-\frac{1}{2}log(x+2)\right]^2_1$

I = 1/2[3log2 – log4 + log3]

I = 1/2[3log2 – 2log2 + log3]

I = 1/2[log 2 – log 3]

I = 1/2[log6]

I = log6/2

Therefore, the value of $\int_{1}^{2}\frac{x+3}{x(x+2)}dx$is log6/2.

问题 38。 $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$

解决方案：

We have,

I = $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$

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

I = $\frac{1}{5}\int_{0}^{1}\frac{10x+15}{5x^2+1}dx$

I = $\frac{1}{5}\int_{0}^{1}(\frac{10x}{5x^2+1}+\frac{15}{5x^2+1}dx$

I = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{1}{5}\int_{0}^{1}\frac{15}{5x^2+1}dx$

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

I = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx$

I = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx$

I = $\frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{5}(\frac{1}{\frac{1}{\sqrt{5}}})tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}\right]^1_0$

I = $\frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0$

I = $\frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0$

I = [1/5log6 + 3/√5tan^-1(√5) – 1/5log1 – 3/√5tan^-1(0)]

I = [1/5 log6 + 3√5 tan^-1(√5) – 0 – 0]

I = 1/5 log6 + 3√5 tan^-1(√5)

Therefore, the value of $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$ is 1/5 log6 + 3√5 tan^-1(√5).

问题 39。 $\int_{0}^{2}\frac{1}{x+4-x^2}dx$

解决方案：

We have,

I = $\int_{0}^{2}\frac{1}{x+4-x^2}dx$

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

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

I = $\int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-\frac{17}{4}}dx$

I = $\int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-(\frac{\sqrt{17}}{2})^2}dx$

I = $\int_{0}^{2}\frac{1}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}dx$

Let x – 1/2 = t, so we have,

=> dx = dt

Now, the lower limit is, x = 0

=> t = x – 1/2

=> t = 0 – 1/2

=> t = 1/2

Also, the upper limit is, x = 2

=> t = x – 1/2

=> t = 2 – 1/2

=> t = 3/2

So, the equation becomes,

I = $\int_{\frac{-1}{2}}^{\frac{3}{2}}\frac{1}{(\frac{\sqrt{17}}{2})^2-t^2}dt$

I = $\left[\frac{1}{2(\frac{\sqrt{17}}{2})}log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]^{\frac{3}{2}}_{\frac{-1}{2}}$

I = $\frac{1}{\sqrt{17}}\left[log\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-log\frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]$

I = $\frac{1}{\sqrt{17}}\left[log\frac{\sqrt{17}+3}{\sqrt{17}-3}-log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]$

I = $\frac{1}{\sqrt{17}}\left[log(\frac{\sqrt{17}+3}{\sqrt{17}-3}×\frac{\sqrt{17}+1}{\sqrt{17}-1})\right]$

I = $\frac{1}{\sqrt{17}}\left[\log\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}}\right]$

I = $\frac{1}{\sqrt{17}}\left[\log\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right]$

I = $\frac{1}{\sqrt{17}}\log\frac{5+\sqrt{17}}{5-\sqrt{17}}$

I = $\frac{1}{\sqrt{17}}\log\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}$

I = $\frac{1}{\sqrt{17}}\log\frac{25+17+10\sqrt{17}}{8}$

I = $\frac{1}{\sqrt{17}}\log\frac{42+10\sqrt{17}}{8}$

I = $\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$

Therefore, the value of $\int_{0}^{2}\frac{1}{x+4-x^2}dx$ is $\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$.

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

解决方案：

We have,

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

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

I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{1}{2}-\frac{1}{16}}dx$

I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{7}{16}}dx$

I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+(\frac{\sqrt{7}}{4})^2}dx$

I = $\left[\frac{1}{2}\frac{4}{\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0$

I = $\left[\frac{4}{2\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0$

I = $\frac{4}{2\sqrt{7}}\left[\tan^{-1}(\frac{\frac{5}{4}}{\frac{\sqrt{7}}{4}})-\tan^{-1}(\frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]$

I = 4/2√7[tan^-1(5/√7) – tan^-1(1/√7)]

I = 2/√7[tan^-1(5/√7) – tan^-1(1/√7)]

Therefore, the value of $\int_{0}^{1}\frac{1}{2x^2+x+1}dx$is 2/√7[tan^-1(5/√7) – tan^-1(1/√7)].

问题 41。 $\int_{0}^{1}\sqrt{x(1-x)}dx$

解决方案：

We have,

I = $\int_{0}^{1}\sqrt{x(1-x)}dx$

Let x = sin² t, so we have,

=> dx = 2 sin t cos t dt

Now, the lower limit is, x = 0

=> sin² t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1

=> sin² t = 1

=> sin t = 1

=> t = π/2

So, the equation becomes,

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

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(cos^2t)}(2sintcost)dt$

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

I = $\int_{0}^{\frac{\pi}{2}}(2sin^2tcos^2t)dt$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(4sin^2tcos^2t)dt$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(sin^22t)dt$

I = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1-cos4t}{2})dt$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos4t)dt$

I = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}}dt-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(cos4t)dt$

I = $\frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{4}\left[\frac{sin4t}{4}\right]^{\frac{\pi}{2}}_0$

I = $\frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{16}\left[sin4t\right]^{\frac{\pi}{2}}_0$

I = 1/4[π/2 – 0] – 1/16[sin2π – 0]

I = 1/4[π/2] – 1/16[0 – 0 ]

I = π/8

Therefore, the value of $\int_{0}^{1}\sqrt{x(1-x)}dx$is π/8.

问题 42。 $\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx$

解决方案：

We have,

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

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

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

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

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

I = [sin^-1(1/2) – sin^-1(-1/2)]

I = π/6 -(-π/6)

I = π/6 + π/6

I = π/3

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

问题 43。 $\int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx$

解决方案：

We have,

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

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

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

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

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

I = $\left[sin^{-1}(\frac{4-2}{2})-sin^{-1}(\frac{0-2}{2})\right]$

I = [sin^-1(2/2) – sin^-1(-2/2)]

I = sin^-11 – sin^-1(-1)

I = π/2 – (-π/2) 

I = π/2 + π/2 

I = π

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

问题 44。 $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$

解决方案：

We have,

I = $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$

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

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

Let x + 1 = t, so we have,

=> dx = dt

Now, the lower limit is, x = –1

=> t = x + 1

=> t = – 1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = $\int_{0}^{2}\frac{1}{t^2+2^2}dt$

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

I = 1/2tan^-12/2 – 1/2tan^-10/2

I = 1/2tan^-11 – 1/2tan^-10

I = 1/2(π/4) – 0

I = π/8

Therefore, the value of $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$is π/8.