第12类RD Sharma解–第19章不定积分–练习19.28

问题1.查找$\int\sqrt{3+2x-x^2}dx$

解决方案：

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

Let considered x – 1 = t, 

so that dx = dt

Thus, $\int\sqrt{3+2x+x^2}dx\\=\int\sqrt{4-t^2}dt\\ =\frac{1}{2}t\sqrt{4-t^2}+\frac{4}{2}sin^{-1}\left(\frac{t}{2}\right)+C\\ =\frac{1}{2}(x-1)\sqrt{3+2x-x^2}+2sin^{-1}\left(\frac{x-1}{2}\right)+C$  

问题2.评估$\int\sqrt{x^2+x+1}dx$

解决方案：

Let I = $\int\sqrt{x^2+x+1}dx\\ =\int\sqrt{x^2+x+\frac{1}{4}+\frac{3}{4}}dx\\ =\int\sqrt{\left(x+\frac{1}{2}^2\right)+\left(\frac{\sqrt{3}}{2}\right)^2}dx\\ =\frac{\left(x+\frac{1}{2}\right)}{2}\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\frac{\left(\frac{\sqrt{3}}{2}\right)^2}{2}.log\left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c\\ =\left(\frac{2x+1}{4}\right)\sqrt{x^2+x+1}+\frac{3}{8}log\left|\left(\frac{2x+1}{2}\right)+\frac{1}{2}\sqrt{x^2+x+1}\right|+c\\ I=\left(\frac{2x+1}{4}\right)\sqrt{x^2+x+1}+\frac{3}{8}log\left|{2x+1}+\sqrt{x^2+x+1}\right|+c\\$

问题3.评估$\int\sqrt{x-x^2}dx$

解决方案：

I = $\int\sqrt{x-x^2}dx\\ \int\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}dx\ \ \ \ \ \ -(Add\ and\ subtract \frac{1}{4})\\ \\ \int\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}-x\right)^2}dx\\ = -\left(\frac{1-2x}{4}\right)\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}-x\right)^2}-\frac{\left(\frac{1}{2}\right)^2}{2}sin^{-1}\left(\frac{\frac{1-x}{2}}{\frac{1}{2}}\right)+c\\$

Hence, $I=\left(\frac{2x-1}{4}\right)\sqrt{x-x^2}+\frac{1}{8}sin^{-1}(2x-1)+c$

问题4：评估$\int\sqrt{1+x-2x^2}dx$

解决方案：

Let I = $\int\sqrt{1+x-2x^2}dx\\ =\sqrt{2}\int\sqrt{\frac{1}{2}+\frac{x}{2}-x^2}dx\\ =\sqrt{2}\int\sqrt{\frac{9}{16}+-\left(\frac{1}{16}-\frac{x}{2}+x\right)^2}dx\\ =\sqrt{2}\int\sqrt{\left(\frac{3}{4}\right)^2-\left(x-\frac{1}{4}\right)^2}dx\\ =\sqrt{2}\left\{\frac{\left(x-\frac{1}{4}\right)}{2}\sqrt{\frac{1}{2}+\frac{x}{2}-x^2}+\frac{9}{32}sin^{-1}\left(\frac{x-\frac{1}{4}}{\frac{3}{4}}\right)\right\}+c\\$

Therefore, I = $\frac{1}{8}(4x-1)\sqrt{1+x-2x^2}+\frac{9\sqrt{2}}{32}sin^{-1}\left(\frac{4x-1}{3}\right)+c$

问题5 $\int cosx\sqrt{4-sin^2x}dx$

解决方案：

I = $\int cosx\sqrt{4-sin^2x}dx$

Let us considered sinx = t

So, on differentiating, we get

cosx dx = dt

I = $\int\sqrt{4-t^2}dt\\ =\int\sqrt{2^2-t^2}dt\\ =\frac{t}{2}\sqrt{2^2-t^2}+\frac{4}{2}sin^{-1}\frac{t}{2}+c\\$

Therefore, I = $\frac{1}{2}sinx\sqrt{4-sin^2x}+2sin^{-1}\left(\frac{sinx}{2}\right)+c$

问题6：评估$\int e^x\sqrt{e^{2x}+1}dx$

解决方案：

I = $\int e^x\sqrt{e^{2x}+1}dx$

Let us considered e^x = t

So, on differentiating, we get

e^xdx = dt

Therefore, I = $\int\sqrt{t^2+1^2}dt\\ =\frac{t}{2}\sqrt{t^2+1^2}+\frac{1}{2}log\left|t+\sqrt{t^2+1}\right|+c$

Hence, I = $\frac{e^x}{2}\sqrt{e^{2x}+1}+\frac{1}{2}log\left|e^x+\sqrt{e^{2x}+1}\right|+c$

问题7：评估$\int\sqrt{9-x^2}dx$

解决方案：

I = $\int\sqrt{3^2-x^2}$

We already have, 

$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}+c$

Therefore, I = $\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}sin^{-1}\frac{x}{3}+c$

问题8.评估$\int\sqrt{16x^2+25}dx$

解决方案：

Let us assume I = $\int\sqrt{16x^2+25}dx$

$=\int\sqrt{(4x)^2+5^2}dx\\ =4\int\sqrt{x^2+\left(\frac{5}{4}\right)^2}dx\\ =4\left\{\frac{x}{2}\sqrt{x^2+\left(\frac{5}{4}\right)^2}+{\frac{\left(\frac{5}{4}\right)^2}{2}}log\left|x+\sqrt{x^2+\left(\frac{5}{4}\right)^2}\right|\right\}+c\\$

Therefore, I = $2x\sqrt{x^2+\frac{25}{16}}+\frac{25}{8}log\left|x+\sqrt{x^2+\frac{25}{16}}\right|+c\\$

问题9.评估$\int\sqrt{4x^2-5}dx$

解决方案：

Let us assume I = $\int\sqrt{4x^2-5}dx$

$=2\int\sqrt{x^2-\left(\frac{\sqrt5}{2}\right)}dx\\ =2\left\{\frac{x}{2}\sqrt{x^2-\frac{5}{4}}-\frac{5}{8}log\left|x+\sqrt{x^2-\frac{5}{4}}\right|+c\right\}$

Therefore, I = $x\sqrt{x^2-\frac{5}{4}}-\frac{5}{4}log\left|x+\sqrt{x^2-\frac{5}{4}}\right|+c$

问题10：评估$\int\sqrt{2x^2+3x+4}dx$

解决方案：

Let us assume I = $\int\sqrt{2x^2+3x+4}dx$

$=\sqrt{2}\int\sqrt{x^2+\frac{3}{2}x+2}dx\\ =\sqrt{2}\int\sqrt{x^2+\frac{3}{2}x+\frac{9}{16}+\frac{23}{16}}dx\\ =\sqrt{2}\int\sqrt{\left(x+\frac{3}{4}\right)^2+\left(\frac{\sqrt{23}}{4}\right)^2}dx\\ =\sqrt{2}\left\{\frac{\left(x+\frac{3}{4}\right)}{2}\sqrt{x^2+\frac{3}{2}x+2}+\frac{23}{32}.log\left|\left(x+\frac{3}{4}\right)+\sqrt{x^2+\frac{3}{2}x+2}\right|+c\right\}$

Therefore, I = $\frac{4x+3}{8}\sqrt{2x^2+3x+4}+\frac{23\sqrt{2}}{32}.log\left|\left(x+\frac{3}{4}\right)+\sqrt{x^2+\frac{3}{2}x+2}\right|+c$

问题11：评估$\int\sqrt{3-2x-2x^2}dx$

解决方案：

Let us assume I = $\int\sqrt{3-2x-2x^2}dx$

$=\sqrt{2}\int\sqrt{\frac{3}{2}-x-x^2}dx\\ =\sqrt{2}\int\sqrt{\frac{7}{4}-\left(\frac{1}{4}+x+x^2\right)}dx\ \ \ \ -(Add\ and\ subtract\ \frac{1}{4})\\ =\sqrt{2}\int\sqrt{\left(\frac{\sqrt{7}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2}dx\\ =\sqrt{2}\left\{\frac{x+\frac{1}{2}}{2}\sqrt{\frac{3}{2}-x+x^2}+\frac{7}{8}sin^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{7}}{2}}\right)+c\right\}$

Therefore, I = $\frac{2x+1}{4}\sqrt{3-2x-2x^2}+\frac{7\sqrt{2}}{8}sin^{-1}\left(\frac{2x+1}{\sqrt{7}}\right)+c$

问题12：评估$\int x\sqrt{x^4+1}dx$

解决方案：

Let us assume x² = t

On differentiating we get

2x dx = dt

Therefore, I = $\frac{1}{2}\int\sqrt{t^2+1^2}dt\\ =\frac{1}{2}\left\{\frac{t}{2}\sqrt{t^2+1}+\frac{1}{2}log\left|t+\sqrt{t^2+1}\right|\right\}+c\\$

Hence, I = $\frac{1}{2}\left\{\frac{x^2}{2}\sqrt{x^4+1}+\frac{1}{2}log\left|x^2+\sqrt{x^4+1}\right|\right\}+c\\$

问题13：评估$\int x^2\sqrt{a^6-x^6}dx$

解决方案：

I = $\int x^2\sqrt{a^6-x^6}dx$

Let us considered x³ = t

So, on differentiating, we get

3x²dx = dt

Therefore, I = $\frac{1}{3}\int\sqrt{a^6-t^2}dt\\ = \frac{1}{3}\left\{\frac{t}{2}\sqrt{a^6-t^2}+\frac{a^6}{2}sin^{-1}\left(\frac{t}{a^3}\right)\right\}+c\\$

Hence, I = $\frac{x^3}{6}\sqrt{a^6-x^6}+\frac{a^6}{6}sin^{-1}\left(\frac{x^3}{a^3}\right)+c$

问题14：评估$\int\sqrt{\frac{16+(logx)^2}{x}}dx$

解决方案：

I = $\int\sqrt{\frac{16+(logx)^2}{x}}dx$

Let us considered logx = t

So, on differentiating, we get

1/x dx = dt 

Therefore, I = $\int\sqrt{16+t^2}dt\\ =\int\sqrt{4^2+t^2}dt\\ =\frac{t}{2}\sqrt{16+t^2}+\frac{16}{2}log\left|t+\sqrt{16+t^2}\right|$

Hence, I = $\frac{logx}{2}\sqrt{16+(logx)^2}+8log\left|logx+\sqrt{16+(logx)^2}\right|+c$

问题15：评估$\int\sqrt{2ax-x^2}dx$

解决方案：

I = $\int\sqrt{2ax-x^2}dx$

$=\int\sqrt{a^2-(a^2-ax+x^2)}dx\ \ \ \ \ -(Add\ and\ subtract\ a^2)\\ =\int\sqrt{a^2-(a-x)^2}dx\\ =\int\sqrt{a^2-(x-a)^2}dx\\ =\frac{(x-a)}{2}\sqrt{2ax-x^2}+\frac{a^2}{2}sin^{-1}\left(\frac{x-a}{a}\right)+c$

Therefore, I = $\frac{1}{2}(x-a)\sqrt{2ax-x^2}+\frac{a^2}{2}sin^{-1}\left(\frac{x-a}{a}\right)+c$

问题16：评估$\int\sqrt{3-x^2}dx$

解决方案：

Let I = $\int\sqrt{3-x^2}dx$

$=\int\sqrt{(\sqrt{3})^2-x^2}dx$

I = $\frac{x}{2}\sqrt{3-x^2}+\frac{3}{2}sin^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$