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.
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