第 12 类 RD Sharma 解决方案 - 第 20 章定积分 - 练习 20.4 A 部分

计算以下每个积分 (1-16)：

问题 1。 $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

解决方案：

We know that $\int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx$

so,

$\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{sin(2π-x)}}{e^{sin(2π-x)}+e^{-sin(2π-x)}}dx$

we know, $sin(2π -x)=-sinx$

$\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$

I = $\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$

then

I = $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

2I = $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx +\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$

$2I=\int\limits_0^{2π}(\frac{e^{sinx}}{e^{sinx}+e^{-sinx}} +\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}})dx$

$2I=\int\limits_0^{2π}(\frac{e^{sinx}+e^{-sinx}}{e^{sinx}+e^{-sinx}} )dx$

$2I=\int\limits_0^{2π}dx$

$2I=2π$

l=π

编程需要懂一点英语

问题2。 $\int\limits_0^{2π} log (secx+tanx)dx$

解决方案：

We know that $\int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx$

So,

$\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (sec(2π-x)+tan(2π-x)dx$

$\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (secx-tanx)dx$

I = $\int\limits_0^{2π} log (secx-tanx)dx$

then

I = $\int\limits_0^{2π} log (secx+tanx)dx$

2I = $\int\limits_0^{2π} log (secx+tanx)dx+ \int\limits_0^{2π} log (secx-tanx)dx$

2I= $\int\limits_0^{2π} log (sec^2x-tan^2x)dx$

2I = $\int\limits_0^{2π} log (1)dx$

2I=0

I=0

编程需要懂一点英语

问题 3。 $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

解决方案：

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

So,

$\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{tan(π/2-x)}}{\sqrt{tan(π/2-x)}+\sqrt{cot(π/2-x)}}dx$

$\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$

I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

then

I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$

2I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$

2I = $\int\limits_{π/6}^{π/3}(\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}} +\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}})dx$

2I= $\int\limits_{π/6}^{π/3}dx$

2I=π/6

I=π/12

编程需要懂一点英语

问题 4。 $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

解决方案：

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

So,

$\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{sin(π/2-x)}}{\sqrt{sin(π/2-x)}+\sqrt{cos(π/2-x)}} dx$

$\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}} dx$

I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

then,

I = $\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

2I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

2I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

2I = $\int\limits_{π/6}^{π/3}dx$

2I=π/6

I=π/12

编程需要懂一点英语

问题 5。 $\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

解决方案：

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

so,

$\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx =\int\limits_{-π/4}^{π/4} \frac{tan^2{-x}}{1+e^{-x}}dx$

$\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$

$I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

then,

$I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{e^xtan^2x}{1+e^x}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x+e^xtan^2x}{1+e^x}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{(e^x+1)tan^2x}{1+e^x}dx$

$2I=\int\limits_{-π/4}^{π/4} tan^2xdx$

$I=\frac{\int\limits_{-π/4}^{π/4} tan^2x}{2}dx$

we know if

f(x) is even $\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx$

f(x) is odd $\int\limits_{-a}^af(x)dx=0$

Here, f(x) = tan²x which is even

hence,

$I=\int\limits_{0}^{π/4}tan^2xdx$

$I=\int\limits_{0}^{π/4}(sec^2x-1)dx$

$I=\left.({tanx-x})\right|_0^{π/4}$

I = $I=π/4$

编程需要懂一点英语

问题 6。 $\int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0$

解决方案：

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

So,

$\int\limits_{-a}^a \frac{1}{1+a^x}dx = \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$

if, $I = \int\limits_{-a}^a \frac{1}{1+a^x}dx$ then $I=\int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$

So,

$2I=\int\limits_{-a}^a \frac{1}{1+a^x}dx+ \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$

$2I=\int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{1}{1+a^{-x}}dx$

$2I= \int\limits_{-a}^a \frac{1+a^x}{1+a^x}dx$

$2I= \int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{a^x}{1+a^x}dx$

$2I= \int\limits_{-a}^a dx$

$2I=2a$

$I=a$

编程需要懂一点英语

问题 7。 $\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

解决方案：

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

Hence,

$\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx =\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$

if, $I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

then $I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$

so,

$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx+\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$

$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{1}{1+e^{-tanx}} dx$

$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{e^{tanx}}{1+e^{tanx}} dx$

$2I=\int\limits_{-π/3}^{π/3} dx$

$2I=2π/3$

$I=π/3$

编程需要懂一点英语

问题 8。 $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

解决方案：

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

hence, $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx =\int\limits_{-π/2}^{π/2} \frac{cos^2(-x)}{1+e^{-x}}dx$

$\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx = \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$

if $I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Then, $I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$

So, $2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx +\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$

$2I= \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}+ \frac{e^xcos^2x}{1+e^x}dx$

$2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x+e^xcos^2x}{1+e^x}dx$

$2I=\int\limits_{-π/2}^{π/2} \frac{(1+e^x)cos^2x}{1+e^x}dx$

$2I=\int\limits_{-π/2}^{π/2} cos^2x dx$

$2I=\int\limits_{-π/2}^{π/2} \frac{1+cos2x}{2}dx$

$I= \left.\frac{1}{4}(x+\frac{sin2x}2)\right|_{-π/2}^{π/2}$

$I=\frac{1}{4}[\frac{π}2-(-\frac{π}2)]$

$I=\frac{π}4$

编程需要懂一点英语

问题 9。 $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

解决方案：

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\frac{1}{cos^2x} dx$

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\int\limits_{-π/4}^{π/4}sec^2x dx$

if f(x) is even

$\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx$

if f(x) is odd

$\int\limits_{-a}^af(x)dx=0$

here, $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}dx$ is odd and $\int\limits_{-π/4}^{π/4}sec^2x dx$

is even

Hence,

$0+2\int\limits_0^{π/4}sec^2x dx$

$\left.2tanx\right|_0^{\frac{π}{4}}$

编程需要懂一点英语

问题 10。 $\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2$

解决方案：

$I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

then,

$I=\int\limits_{a}^{b} \frac{{(a+b-x)}^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

$2I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx +\int\limits_{a}^{b} \frac{(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

$2I=\int\limits_{a}^{b} \frac{x^{1/n}+(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

$2I=\int\limits_{a}^{b} dx$

$I=\frac{b-a}{2}$

编程需要懂一点英语

问题 11。 $\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

解决方案：

let, $I=\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

$I=\int\limits_0^{π/2} (log\frac{cos^2x}{sinx2x})dx$

$I=\int\limits_0^{π/2} (log\frac{cos^2x}{2sinxcosx})dx$

$I=\int\limits_0^{π/2} (log\frac{cosx}{2sinx})dx$

$I=\int\limits_0^{π/2} logcosxdx-\int\limits_0^{π/2} logsinxdx-\int\limits_0^{π/2}log2dx$

we know that, $\int\limits_0^{π/2}logcosxdx=\int\limits_0^{π/2}logsinxdx$

hence,

$I=-\int\limits_0^{π/2}log2dx =\frac{-π}{2}log2$

编程需要懂一点英语

问题 12。 $\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

解决方案：

Let, $I=\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

we know that , $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$

so, $I=\int\limits_0^a \frac{\sqrt {a-x}}{\sqrt{a-x}+\sqrt {x}} dx$

then,

$2I= \int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx + \int\limits_0^a \frac{\sqrt{ a-x}}{\sqrt {a-x}+\sqrt x} dx$

$2I= \int\limits_0^a \frac{\sqrt x+\sqrt{a-x}}{\sqrt x+\sqrt {a-x}} dx$

$2I= \int\limits_0^a dx$

$2I=a$

$I=\frac{a}{2}$

编程需要懂一点英语

问题 13。 $\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

解决方案：

We know that, $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$

So,

$I=\int\limits_0^5 \frac {\sqrt[4]{(5-x)+4} }{\sqrt[4]{(5-x)+4} +\sqrt[4]{9-(5-x)} }dx$

$I=\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx$

then,

$2I=\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx +\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx$

$2I= \int\limits_0^5 \frac{\sqrt[4]{x+4}\sqrt[4]{9-x} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

$2I= \int\limits_0^5 dx$

I $I=\frac{5}{2}$

编程需要懂一点英语

问题 14。 $\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx$

解决方案：

We know that, $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$

so,

$I=\int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx$

then,

$2I=\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx+ \int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx$

$2I=\int\limits_0^7 dx$

$I=\frac{7}{2}$

编程需要懂一点英语

问题 15。 $\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

解决方案：

We know that, $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

Let, $I=\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

hence,

$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cos(\frac{π}{2}- x)}}{\sqrt{cos(\frac{π}{2}-x)} +\sqrt {sin(\frac{π}{2}-x)}}dx$

$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

$2I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx + \int\limits_{π/6}^{π/3} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

$2I= \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}+\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

$2I= \int\limits_{π/6}^{π/3} dx$

$I=\frac{π}{12}$

编程需要懂一点英语

问题 16. 如果 f(a+bx)=f(x)，则证明 $\int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx$

解决方案：

$I=\int\limits_a^bf(x)dx$

$I=\int\limits_a^b(a+b-x)f(a+b-x)dx$

$I=\int\limits_a^b(a+b-x)f(x)dx........... [\because f(a+b-x)=f(x)]$

$I=\int\limits_a^b(a+b)f(x)dx-\int\limits_a^bf(x)dx$

$I=(a+b)\int\limits_a^bf(x)dx-I$

$2I=(a+b)\int\limits_a^bf(x)dx$

$I=\frac{(a+b)}{2}\int\limits_a^bf(x)dx$

$\therefore \int\limits_a^bxf(x)dx=\frac{(a+b)}{2}\int\limits_a^bf(x)dx$

编程需要懂一点英语