三角替代

任何人都可以通过替代集成是一种很好且最简单的方法。当我们替换一个函数时使用它，该函数的派生已包含在给定的积分函数。这样，函数得到简化，并且获得了易于积分的简单积分函数。这也称为u替代或反向链规则。换句话说，使用这种方法，我们可以轻松地评估积分和反导数。

The integration by substitution method:

Let us take an integral I = ∫f(x)dx  …..(i)

Now we can transform this integral by changing its independent variable x to u by substituting x = g(u).

Differentiate g(u) w.r.t.x, we get

dx/du = g'(u)

or we can write as

dx = g'(u) du

Now put all these values in eq(i), we get

∫f(x)dx = ∫f(g(u)).g'(u) du

因此，替代整合的一般形式是

∫f(g(x)).g′(x).dx = ∫f(u).dt

这里u = g(x)。替代积分有两种类型：

1.直接替代：直接替代 代换 遵循替换定理的积分。

Theorem:

Let us assume f be the function defined in I interval and has an inverse function F. Let g be the function from H interval into I interval which is differentiable on H. So, F(g) is an inverse function or antiderivative of f(g)g’ on H:

∫f(g(x)).g′(x).dx = ∫f(y).dy\_{y = g(x)}

在直接替换中，如果给定的积分形式为∫f(g(x))。g′(x).dx(通过重新排列)。然后，我们可以通过代入u = g(x)将其独立变量x更改为u来转换此积分。

∫f(g(x))。g′(x).dx =∫f(u).g′(x).dx

现在我们必须用du替换g’(x)dx。

所以我们得到

∫f(g(x))。g′(x).dx =∫f(u).du

现在最终结果是这种形式

F(u)+ C = F(g(x))+ C

Example: ∫sin(x²)8xdx

As, here it is 8x not 2x. 

Then we will try to make it fit into the definition. We create its derivative here and do some rearranging of functions.

= ∫4sin(x²)2xdx

= 4 $\int sin(u)\hspace{0.1cm}du$

= 4 (- cos(u) + c)

= – 4 cos(u) + c₁  (c₁ = 4c)

Now, substituting u = x^2, we get

∫sin(x²)8xdx = – 4 cos(x²) + c₁

2.间接替换：这也是一种包括简化的替换方法，但与直接替换相反。

在间接替换中，如果给定的积分形式为∫f(u).du。然后，我们可以通过替换u = g(x)将该积分转换回其原始自变量u到x。

∫F(U).du =∫F(G(X))杜 

现在我们必须用g’(x)dx替换du。

所以我们得到

∫F(G(X))杜 =∫F(G(X))G’(x)的DX

现在最终结果是这种形式

G(u)+ C = G(g ^-1 (x))+ C

如前所述，看起来好像我们在相反的方向上遵循了上面的直接替换。间接替换很少使用，但是有特定类型的积分。通常，在与根进行积分的地方很有用。

Example: $\int x^2 \sqrt{x^3+5} \hspace{0.1cm}dx$

Let’s substitute

u = $\sqrt{x^3+5}$

u² = x³ + 5

Differentiating, w.r.t to x, we get

2u du/dx = 3x²

x² dx = 2/3u du

By substituting these values, we get

$\int u \hspace{0.1cm}(\frac{2}{3}u du)$

$\frac{2}{3} \int u^2 \hspace{0.1cm}du$

$\frac{2}{3} (\frac{u^3}{3} + c)$

2/9 u³+ c₁ (c₁ = 2/9 c)

As, u = $\sqrt{x^3+5}$, we get 

I = $\frac{2}{9} (\sqrt{x^3+5})^3$ + c₁

三角替代

三角函数取代是一种方法，其中替代吨rigonometric函数到另一表达发生。它用于评估积分，或者是一种查找包含二次表达式的平方根或形式的有理次幂的函数的反导数的方法$\frac{p}{2}$ (其中p是整数)的二次表达式。这样的表达的例子是

$({x^2+4})^\frac{3}{2}$或者$\sqrt{25-x^2}$或其他……

当其他更常见且更易于使用的积分方法失败时，可以调用三角替换法。三角替换假定您熟悉标准三角标识，微分符号的使用，使用u替代的积分以及三角函数的积分。

x = f( θ )

dx = f’( θ ) dθ

在这里，我们将根据需要集成的函数讨论一些重要的公式，我们将替换以下三角表达式之一以简化集成：

∫cosx dx = sinx + C

∫sinx dx = −cosx + C

∫sec²x dx = tanx + C

∫cosec²x dx = −cotx + C

∫secx tanx dx = secx + C

∫cosecx cotx dx = −cosecx + C

∫tanx dx = ln|secx| + C

∫cotx dx = ln|sinx| + C

∫secx dx = ln|secx + tanx| + C

∫cosecx dx = ln|cosecx − cotx| + C
 

以下是一些可用于评估积分的替代方法：

Expression	Substitution
a2 + x2	x = a tan θ or  x = a cot θ
a2 – x2	x = a sin θ or  x = a cos θ
x2 – a2	x = a sec θ or x = a cosec θ
$\sqrt{\frac{a-x}{a+x}}$ or, $\sqrt{\frac{a+x}{a-x}}$	x = a cos 2θ
$\sqrt{\frac{x-\alpha}{\beta-x}}$ or, $\sqrt{(x-\alpha)(x-\beta)}$	x = ∝ cos2θ + β sin2θ

由² – x ²组成的整数。

Example: $\int \frac{1}{\sqrt{a^2-x^2}}\hspace{0.1cm}dx$

Lets put, x = a sinθ

dx = a cosθ dθ

I = $\int \frac{a\hspace{0.1cm}cos\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2-(a\hspace{0.1cm}sin\theta)^2)}}$

I = $\int \frac{a\hspace{0.1cm}cos\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2cos^2\theta)}}$

I = $\int 1. d\theta$

I = θ + c

As, x = a sinθ

⇒ θ = $sin^{-1}(\frac{x}{a})$

I = $sin^{-1}(\frac{x}{a}) + c$

由x ² + a ²组成的整数。

Example: $\int \frac{1}{x^2+a^2}\hspace{0.1cm}dx$

Lets put x = a tanθ

dx = a sec2θ dθ, we get

I = $\int \frac{1}{(a\hspace{0.1cm}tan\theta)^2+a^2}\hspace{0.1cm}(a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta)$

I = $\int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{a^2(sec^2\theta)}$

I = $\frac{1}{a}\int 1.d\theta$

I = $\frac{1}{a} \theta$+ c

As, x = a tanθ

⇒ θ = $tan^{-1}(\frac{x}{a})$

I = $\frac{1}{a}tan^{-1}(\frac{x}{a})$+ c

由² + x ²组成的整数。

Example: $\int \frac{1}{\sqrt{a^2+x^2}}\hspace{0.1cm}dx$

Lets put, x = a tanθ

dx = a sec²θ dθ

I = $\int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2+(a\hspace{0.1cm}tan\theta)^2)}}$

I = $\int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2\hspace{0.1cm}sec^2\theta)}}$

I = $\int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{a\hspace{0.1cm}sec\theta}$

I = $\int sec\hspace{0.1cm}\theta d\theta$

I = $log|sec\hspace{0.1cm}\theta+tan\hspace{0.1cm}\theta| + c$

I = $log|tan\hspace{0.1cm}\theta+\sqrt{1+tan^2\hspace{0.1cm}\theta}| + c$

I = $log|\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}}|+ c$

I = $log|\frac{x}{a}+\sqrt{\frac{a^2+x^2}{a^2}}|+ c$

I = $log|\frac{x}{a}+\frac{1}{{a}}\sqrt{a^2+x^2}|+ c$

I = $log|x+\sqrt{a^2+x^2}|-log\hspace{0.1cm}a+ c$

I = $log|x+\sqrt{a^2+x^2}|+ c_1$

由x ² – a ²组成的整数。

Example: $\int \frac{1}{\sqrt{x^2-a^2}}\hspace{0.1cm}dx$

Let’s put, x = a secθ

dx = a secθ tanθ dθ

I = $\int \frac{a\hspace{0.1cm}sec\theta \hspace{0.1cm}tan\theta\hspace{0.1cm}d\theta}{\sqrt{((a\hspace{0.1cm}sec\theta)^2-a^2)}}$

I = $\int \frac{a\hspace{0.1cm}sec\theta \hspace{0.1cm}tan\theta\hspace{0.1cm}d\theta}{(a\hspace{0.1cm}tan\theta)}$

I = $\int sec\theta\hspace{0.1cm}d\theta$

I = $log|sec\hspace{0.1cm}\theta+tan\hspace{0.1cm}\theta| + c$

I = $log|sec\hspace{0.1cm}\theta+\sqrt{sec^2\hspace{0.1cm}\theta-1}| + c$

I = $log|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}|+ c$

I = $log|\frac{x}{a}+\sqrt{\frac{x^2-a^2}{a^2}}|+ c$

I = $log|\frac{x}{a}+\frac{1}{{a}}\sqrt{x^2-a^2}|+ c$

I =$log|x+\sqrt{x^2-a^2}|-log\hspace{0.1cm}a+ c$

I = $log|x+\sqrt{x^2-a^2}|+ c_1$

样本问题

问题1 $\int \frac{1}{\sqrt{9-25x^2}} \hspace{0.1cm}dx$

解决方案：

Taking 5 common in denominator,

I = $\frac{1}{5}\int \frac{1}{\sqrt{\frac{9}{25}-x^2}} \hspace{0.1 cm} dx$

I = $\frac{1}{5}\int \frac{1}{\sqrt{(\frac{3}{5})^2-x^2}} \hspace{0.1 cm} dx$

According to theorem 1, a = $\frac{3}{5}$

I = $\frac{1}{5} sin^{-1}(\frac{x}{\frac{3}{5}})$ + c

I = $\frac{1}{5} sin^{-1}(\frac{5x}{3})$ + c

问题2。 $\int \frac{1}{\sqrt{8-2x^2}} \hspace{0.1cm}dx$

解决方案：

Taking √2 common in denominator,

I =$\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{\frac{8}{2}-x^2}} \hspace{0.1 cm} dx$

I = $\frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{(2)^2-x^2}} \hspace{0.1 cm} dx$

According to theorem 1, a = 2

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

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

问题3。 $\int x^3\sqrt{9-x^2}\hspace{0.1cm}dx$

解决方案：

By rearranging, we get

$\int x^3\sqrt{3^2-x^2}\hspace{0.1cm}dx$

Here taking,

a = 3

x = 3 sinθ

dx = 3 cos θ dθ

Substituting these values,

I = $\int (3 sinθ)^3\sqrt{(3^2-(3 sin\theta)^2)}\hspace{0.1cm}3 \hspace{0.1cm}cos \theta\hspace{0.1cm}d\theta$

I = $\int 27 sin^3\theta \hspace{0.1cm}3\sqrt{(1-sin^2\theta)}\hspace{0.1cm}3 \hspace{0.1cm}cos \theta\hspace{0.1cm}d\theta$

I = $\int 243 \hspace{0.1cm}sin^3\theta cos^2\theta\hspace{0.1cm}d\theta$

I = 243 $\int\hspace{0.1cm}sin^2\theta \hspace{0.1cm}sin\theta\hspace{0.1cm}cos^2\theta\hspace{0.1cm}d\theta$

I = 243 $\int\hspace{0.1cm}(1-cos^2\theta) \hspace{0.1cm}sin\theta\hspace{0.1cm}cos^2\theta\hspace{0.1cm}d\theta$

Lets take,

u = cos θ

du = -sin θ dθ

Substituting these values, we get

I = 243 $\int\hspace{0.1cm}(1-u^2) \hspace{0.1cm}u^2\hspace{0.1cm}(-du)$

I = -243 $\int\hspace{0.1cm}(u^2-u^4) \hspace{0.1cm}du$

I = -243 $\int\hspace{0.1cm}u^2 \hspace{0.1cm}du - \int\hspace{0.1cm}u^4 \hspace{0.1cm}du$

I = -243 $[\frac{u^3}{3} - \frac{u^5}{5}]$

As, u = cos θ and x = 3 sinθ

cos θ = $\sqrt{1-sin^2\theta}$

u = $\sqrt{1-(\frac{x}{3})^2}$

u = $(1-\frac{x^2}{9})^{\frac{1}{2}}$

Hence,

I =  -243 $[\frac{({(1-\frac{x^2}{9})^{\frac{1}{2}})}^3}{3}-\frac{({(1-\frac{x^2}{9})^{\frac{1}{2}})}^5}{5}]$

I = -243$[\frac{(1-\frac{x^2}{9})^{\frac{3}{2}}}{3}-\frac{(1-\frac{x^2}{9})^{\frac{5}{2}}}{5}]$ + c

问题4。 $\int \frac{1}{4+9x^2} \hspace{0.1cm}dx$

解决方案：

Taking 9 common in denominator,

I = $\frac{1}{9}\int \frac{1}{\frac{4}{9}+x^2} \hspace{0.1 cm} dx$

I = $\frac{1}{9}\int \frac{1}{(\frac{2}{3})^2+x^2} \hspace{0.1 cm} dx$

According to theorem 2, a = $\frac{2}{3}$

I = $\frac{1}{9} \times \frac{1}{\frac{2}{3}}tan^{-1} \frac{x}{(\frac{2}{3})}$

I = $\frac{1}{6}tan^{-1} (\frac{3x}{2})+ c$

问题5 $\int \frac{1}{\sqrt{16x^2+25}}\hspace{0.1cm}dx$

解决方案：

Taking 4 common in denominator,

I = $\frac{1}{4}\int\frac{1}{\sqrt{x^2+\frac{25}{16}}}$

I = $\frac{1}{4}\int\frac{1}{\sqrt{x^2+(\frac{5}{4})^2}}$

According to theorem 3, a = $\frac{5}{4}$

I = $\frac{1}{4}\times log|x+\sqrt{(\frac{5}{4})^2+x^2}|+ c$

I = $\frac{1}{4}\times log|\frac{4x+\sqrt{25+16x^2}}{4}|+ c$

I = $\frac{1}{4}log|4x+\sqrt{25+16x^2}|-\frac{1}{4}log4+ c$

I = $\frac{1}{4}log|4x+\sqrt{25+16x^2}|+ c_1$

问题6。 $\int \frac{1}{\sqrt{4x^2-9}}\hspace{0.1cm}dx$

解决方案：

Taking 2 common in denominator,

I = $\frac{1}{2}\int \frac{1}{\sqrt{x^2-\frac{9}{4}}} \hspace{0.1cm}dx$

I = $\frac{1}{2}\int \frac{1}{\sqrt{x^2-(\frac{3}{2})^2}} \hspace{0.1cm}dx$

According to theorem 4, a = $\frac{3}{2}$

I = $\frac{1}{2}\times log|x+\sqrt{x^2-(\frac{3}{2})^2}|+c$

I = $\frac{1}{2}log|x+\sqrt{x^2-\frac{9}{4}}|+c$

I = $\frac{1}{2}log|\frac{2x+\sqrt{x^2-9}}{2}|+c$

I = $\frac{1}{2}log|2x+\sqrt{x^2-9}|-\frac{1}{2}log2+c$

I = $\frac{1}{2}log|2x+\sqrt{x^2-9}|+c_1$

问题7。 $\int \frac{1}{x^2-x+1}\hspace{0.1cm}dx$

解决方案：

After rearranging, we get

I = $\int \frac{1}{x^2-x+\frac{1}{4}-\frac{1}{4}+1}\hspace{0.1cm}dx$

I = $\int \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4})}\hspace{0.1cm}dx$

I = $\int \frac{1}{(x-\frac{1}{2})^2+(\sqrt{\frac{3}{4}})^2})\hspace{0.1cm}dx$

I = $\int \frac{1}{(x-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2})\hspace{0.1cm}dx$

According to the theorem 2, we have

x = x-$\frac{1}{2}$and a = $\frac{\sqrt{3}}{2}$

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

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