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It is known that in the year 2015, Thailand will be a member of AEC (ASEAN Economic Community). Its main objectives include the creation of a single market and production base, a highly competitive economic region, a region of equitable economic development, and a region fully integrated into the global economy (Wikipedia, 2013)(http://en.wikipedia.org/wiki/ASEAN#From_CEPT_to_AEC). Thai government is concerned about its people and especially Thai youths who cannot communicate in English. Now, English has become the international language of communication for all neighboring countries. As a result, this website is encouraged all Thai students, teachers and parents to get familiarity with English communication in Mathematics which is the important area for students to study in higher levels. |
วันจันทร์ที่ 8 มิถุนายน พ.ศ. 2558
วันเสาร์ที่ 6 มิถุนายน พ.ศ. 2558
Formula for the Differentiation and Integration of Inverse of Trigonometric Functions
1. ARCSINE
1.1 Derivative: \[ \frac{d}{dx}(arcsin(x)) = \frac{1}{\sqrt{1 - x^2}} \]
1.2 integration: \[ \int \frac{1}{\sqrt{1 - x^2}}~dx = arcsin(x) + C \]
Proof. Since \[ y = arcsin (x) \]
It implies that \[ sin (y) = x, x ∈ [\frac{-π}{2}, \frac{π}{2}] \]
Please see the graphs and right triangle for helping calculation below:
From \[ sin(y) = x \]
Differentiate with respect to x both sides \[ \frac{d}{dx}(sin(y)) = \frac{d}{dx}x \]
Then \[ cos(y) \frac{dy}{dx} = 1 \]
Therefore, \[ \frac{dy}{dx} = \frac{1}{cos(y)} \]
\[ = \frac{1}{\frac{\sqrt{1 - x^2}}{1}} \]
\[ = \frac{1}{\sqrt{1 - x^2}} \]
and \[ \int \frac{1}{\sqrt{1 - x^2}}~dx = arcsin(x) + C \]
2. ARCTANGENT
2.1 Derivative: \[ \frac{d}{dx}(arctan(x)) = \frac{1}{1 + x^2} \]
2.2 integration: \[ \int \frac{1}{1 + x^2}~dx = arctan(x) + C \]
Proof. Since \[ y = arctan (x) \]
It implies that \[ tan (y) = x, x ∈ (\frac{-π}{2}, \frac{π}{2}) \]
Please see the graphs and right triangle for helping calculation below:
From \[ tan(y) = x \]
Differentiate with respect to x both sides \[ \frac{d}{dx}(tan(y)) = \frac{d}{dx}x \]
Then \[ sec^2(y) \frac{dy}{dx} = 1 \]
Therefore, \[ \frac{dy}{dx} = \frac{1}{sec^2(y)} \]
\[ = \frac{1}{(\frac{\sqrt{1 + x^2}}{1})^2} \]
\[ = \frac{1}{1 + x^2} \]
and \[ \int \frac{1}{1 + x^2}~dx = arctan(x) + C \]
3. ARCCOSINE
3.1 Derivative: \[ \frac{d}{dx}(arccos(x)) = -\frac{1}{\sqrt{1 - x^2}} \]
3.2 integration: \[ -\int \frac{1}{\sqrt{1 - x^2}}~dx = arccos(x) + C \]
Proof. Since \[ y = arccos (x) \]
It implies that \[ cos (y) = x, x ∈ [0, π] \]
Please see the graphs and right triangle for helping calculation below:
From \[ cos(y) = x \]
Differentiate with respect to x both sides \[ \frac{d}{dx}(cos(y)) = \frac{d}{dx}x \]
Then \[ - sin(y) \frac{dy}{dx} = 1 \]
Therefore, \[ \frac{dy}{dx} = -\frac{1}{sin(y)} \]
\[ = -\frac{1}{\frac{\sqrt{1 - x^2}}{1}} \]
\[ = -\frac{1}{\sqrt{1 - x^2}} \]
and \[ -\int \frac{1}{\sqrt{1 - x^2}}~dx = arccos(x) + C \]
4. ARCCOTANGENT
4.1 Derivative: \[ \frac{d}{dx}(arccot(x)) = -\frac{1}{1 + x^2} \]
4.2 integration: \[ -\int \frac{1}{1 + x^2}~dx = arccot(x) + C \]
Proof. Since \[ y = arccotn (x) \]
It implies that \[ tan (y) = x, x ∈ (0, π) \]
Please see the graphs and right triangle for helping calculation below:
From \[ cot(y) = x \]
Differentiate with respect to x both sides \[ \frac{d}{dx}(cot(y)) = \frac{d}{dx}x \]
Then \[ - cosec^2(y) \frac{dy}{dx} = 1 \]
Therefore, \[ \frac{dy}{dx} = -\frac{1}{cosec^2(y)} \]
\[ = -\frac{1}{(\frac{\sqrt{1 + x^2}}{1})^2} \]
\[ = -\frac{1}{1 + x^2} \]
and \[ -\int \frac{1}{1 + x^2}~dx = arccot(x) + C \]
5. ARCSECANT
5.1 Derivative: \[ \frac{d}{dx}(arcsec(x)) = \frac{1}{x\sqrt{x^2 - 1}} \]
5.2 integration: \[ \int \frac{1}{x\sqrt{x^2 - 1}}~dx = arcsec(x) + C \]
Proof. Since \[ y = arcsec (x) \]
It implies that \[ sec (y) = x, x ∈ [0, \frac{π}{2}) or (\frac{π}{2}, π] \]
Please see the graphs and right triangle for helping calculation below:
From \[ sec(y) = x \]
Differentiate with respect to x both sides \[ \frac{d}{dx}(sec(y)) = \frac{d}{dx}x \]
Then \[ sec(y) tan(y) \frac{dy}{dx} = 1 \]
Therefore, \[ \frac{dy}{dx} = \frac{1}{sec(y)tan(y)} \]
\[ = \frac{1}{\frac{x}{1}\frac{\sqrt{x^2 - 1}}{1}} \]
\[ = \frac{1}{x\sqrt{x^2 - 1}} \]
and \[ \int \frac{1}{x\sqrt{x^2 - 1}}~dx = arcsec(x) + C \]
6. ARCCOSECCANT
6.1 Derivative: \[ \frac{d}{dx}(arccosec(x)) = -\frac{1}{x\sqrt{x^2 - 1}} \]
6.2 integration: \[ -\int \frac{1}{x \sqrt{x^2 - 1}}~dx = arccosec(x) + C \]
Proof. Since \[ y = arccosec (x) \]
It implies that \[ cosec (y) = x, x ∈ [-\frac{π}{2}, 0) or (0, \frac{π}{2}] \]
Please see the graphs and right triangle for helping calculation below:
From \[ cosec(y) = x \]
Differentiate with respect to x both sides \[ \frac{d}{dx}(sec(y)) = \frac{d}{dx}x \]
Then \[ - cosec(y) cot(y) \frac{dy}{dx} = 1 \]
Therefore, \[ \frac{dy}{dx} = - \frac{1}{cosec(y)cot(y)} \]
\[ = -\frac{1}{\frac{x}{1}\frac{\sqrt{x^2 - 1}}{1}} \]
\[ = -\frac{1}{x\sqrt{x^2 - 1}} \]
and \[ -\int \frac{1}{x\sqrt{x^2 - 1}}~dx = arccosec(x) + C \]
CONCLUSION
Within 6 Formula above, we can remember on 3 Formula these are ARCSINE, ARCTANGENT, and ARCSECCANT. For the rest, it has negative sign of the one previously. These mean: ARCCOSINE is the negative sign of ARCSINE, ARCCOTANGENT is the negative sign of ARCTANGENT, and ARCCOSECCANT is the negative sign of ARCSECCANT.
Example 1. Find the derivative of
\[ f(x) = arcsin(3x^2) \]
Solution. Apply the Chain Rule to formula of the derivative of acrsine:
Then \[ f^{'}(x) = \frac{1}{\sqrt{1 - (3x^2)^2}} \frac{d}{dx}(3x^2) \]
\[ = \frac{6x}{\sqrt{1 - 9x^4}} \]
Example 2. Find the derivative of
\[ f(x) = arctan(ax^2 + bx + c) \]
Solution. Apply the Chain Rule to formula of the derivative of acrtangent:
Then \[ f^{'}(x) = \frac{1}{ax^2 + bx + c)^2} \frac{d}{dx}(ax^2 + bx + c) \]
\[ = \frac{2ax + b}{(1 + (ax^2 + bx + c)^2} \]
Example 3. Prove that:
\[ \int \frac{1}{\sqrt{a^2 - x^2}}~dx = arcsin\frac{x}{a} + C, a>0 \]
Proof. Apply Substitution Rule to arcsin:
Let \[ x = au, u = \frac{x}{a} \]
Then \[ dx = adu \]
It implies that \[ \int \frac{1}{\sqrt{a^2 - x^2}}~dx = \int \frac{1}{\sqrt{a^2 - a^2u^2}}~(adx) \]
\[ = \int \frac{1}{\sqrt{a^2(1 - u^2)}}~(adx) \]
\[ = \int \frac{1}{a\sqrt{1 - u^2}}~(adx) \]
\[ = \int \frac{1}{\sqrt{1 - u^2}}~dx \]
\[ = arcsin(u) + C \]
\[ = arcsin\frac{x}{a} + C \]
Example 4. Prove that:
\[ \int \frac{1}{a^2 - x^2}~dx = \frac{1}{a} arctan\frac{x}{a} + C, a>0 \]
Proof. Apply Substitution Rule to arctan:
Let \[ x = au, u = \frac{x}{a} \]
Then \[ dx = adu \]
It implies that \[ \int \frac{1}{a^2 - x^2}~dx = \int \frac{1}{a^2 - a^2u^2}~(adx) \]
\[ = \frac{1}{a^2} \int \frac{1}{1 - u^2}~(adx) \]
\[ = \frac{1}{a} \int \frac{1}{1 - u^2}~dx \]
\[ = \frac{1}{a} arctan(u) + C \]
\[ = \frac{1}{a} arctan\frac{x}{a} + C \]
Direction:
1. Choose the appropriate answer to the derivative of trigonometric questions
2. After finished 10 questions, Click the button of "Check for All Questions"
3. For wrong answer, Click the button of "Show Answer" for seeing its correct answer
Question 1:Which one of the following is equal to the derivative of:
\[ y = arctan (x + 1) \]
1. \[ \frac{dy}{dx} = -\frac{1}{x^2 + 2x + 2} \]
2. \[ \frac{dy}{dx} = -\frac{1}{x^2 + 2x + 4} \]
3. \[ \frac{dy}{dx} = \frac{1}{x^2 + 2x + 4} \]
4. \[ \frac{dy}{dx} = \frac{1}{x^2 + 2x + 2} \]
Question 2:Which one of the following is equal to the derivative of:
\[ y = arcsin (x^2) \]
1. \[ \frac{dy}{dx} = -\frac{2x}{\sqrt{1 - x^2}} \]
2. \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]
3. \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]
4. \[ \frac{dy}{dx} = \frac{2x}{\sqrt{1 - x^2}} \]
Question 3:Which one of the following is equal to the derivative of:
\[ y = (x)arcsin (2x) \]
1. \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^4}} + arcsin(2x) \]
2. \[ \frac{dy}{dx} = \frac{2x}{\sqrt{1 - x^4}} + arcsin(2x) \]
3. \[ \frac{dy}{dx} = -\frac{2x}{\sqrt{1 - x^4}} + arcsin(2x) \]
4. \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^4}} - arcsin(2x) \]
Question 4:Which one of the following is equal to the derivative of:
\[ y = (arcsin(x))^2 \]
1. \[ \frac{dy}{dx} = -\frac{2arccos(x)}{\sqrt{1 - x^2}} \]
2. \[ \frac{dy}{dx} = \frac{2arccos(x)}{\sqrt{1 - x^2}} \]
3. \[ \frac{dy}{dx} = \frac{2arcsin(x)}{\sqrt{1 - x^2}} \]
4. \[ \frac{dy}{dx} = -\frac{2arcsin(x)}{\sqrt{1 - x^2}} \]
Question 5:Which one of the following is equal to the derivative of:
\[ y = arcsin \frac{c + r}{1 - cr}, c >0 \]
1. \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 + r^2}} \]
2. \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - r^2}} \]
3. \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - r^2}} \]
4. \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 + r^2}} \]
Question 6:Which one of the following is equal to the derivative of:
\[ θ = arcsin\sqrt{1 - r^2} \]
1. \[ \frac{dθ}{dr} = \frac{r}{|r|\sqrt{1 + r^2}} \]
2. \[ \frac{dθ}{dr} = \frac{r}{|r|\sqrt{1 - r^2}} \]
3. \[ \frac{dθ}{dr} = -\frac{r}{|r|\sqrt{1 + r^2}} \]
4. \[ \frac{dθ}{dr} = -\frac{r}{|r|\sqrt{1 - r^2}} \]
Question 7:Which one of the following is equal to the derivative of:
\[ f(x) = \sqrt{c - x^2} + (c)arcsin\frac{x}{c}, c>0 \]
1. \[ f^{'}(x) = \sqrt{\frac{c-x}{c+x}} \]
2. \[ f^{'}(x) = \sqrt{\frac{c+x}{c+x}} \]
3. \[ f^{'}(x) = \sqrt{\frac{c-x}{c-x}} \]
4. \[ f^{'}(x) = \sqrt{\frac{c+x}{c-x}} \]
Question 8:Which one of the following is equal to the value of:
\[ \int_{-1}^1 \frac{1}{1 + x^2}~dx \]
1. \[ Answer = \frac{π}{8} \]
2. \[ Answer = \frac{π}{6} \]
3. \[ Answer = \frac{π}{4} \]
4. \[ Answer = \frac{π}{2} \]
Question 9:Which one of the following is equal to the value of:
\[ \int_0^1 \frac{1}{\sqrt{4 - x^2}}~dx \]
1. \[ Answer = \frac{π}{2} \]
2. \[ Answer = \frac{π}{4} \]
3. \[ Answer = \frac{π}{6} \]
4. \[ Answer = \frac{π}{8} \]
Question 10:Which one of the following is equal to the value of:
\[ \int_0^1 \frac{1}{1 + x^2}~dx \]
1. \[ Answer = \frac{π}{2} \]
2. \[ Answer = \frac{π}{4} \]
3. \[ Answer = \frac{π}{6} \]
4. \[ Answer = \frac{π}{8} \]