Integral of Arctan (Tan Inverse x)
The integral of arctan is the integration of tan inverse x, which is also called the antiderivative of arctan, which is given by ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C, where C is the constant of integration. The integral of arctan can be calculated using the integration by parts method. Integration is the process of reverse differentiation, that is, determining the antiderivative.
Let us learn the integration of tan inverse x, prove that the arctan integral is ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C and determine the definite integral of arctan along with some solved examples for a better understanding.
1. | What is the Integral of Arctan? |
2. | Integral of Arctan Proof Using Integration by Parts |
3. | Definite Integral of Tan Inverse x From 0 to 1 |
4. | FAQs on Integral of Arctan |
What is the Integral of Arctan?
The integral of arctan also called as integral of tan inverse x, is x tan-1x - ½ ln |1+x2| + C. Mathematically, it is written as ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C. Here, C is the constant of integration, dx denotes that the integration of tan inverse x is with respect to x, and ∫ denotes the symbol of integration. The integral of arctan can be calculated using the ILATE method, that is, integration by parts.
Integral of Arctan Formula
The formula for integral of arctan is given by,
Integral of Arctan Proof Using Integration by Parts
Now that we know that the integral of arctan is x tan-1x - ½ ln |1+x2| + C, we will prove it using the integration by parts method. We will use the following formulas and facts to prove the integral of arctan.
- The formula for integration by parts is ∫f(x)g(x)dx = f(x) ∫g(x)dx - ∫[d(f(x))/dx × ∫g(x) dx] dx.
- Note that tan-1x can be written as tan-1x = tan-1x.1
- We have f(x) = tan-1x, g(x) = 1
- d(tan-1x)/dx = 1/(1 + x2)
Using these formulas and facts, we have
∫tan-1x dx = ∫tan-1x.1 dx
= tan-1x ∫1dx - ∫[d(tan-1x)/dx × ∫1 dx] dx
= x tan-1x - ∫[1/(1 + x2) × x] dx
= x tan-1x - ∫x/(1 + x2) dx
= x tan-1x - (1/2) ∫2x/(1 + x2) dx [Multiplying and dividing by 2]
= x tan-1x - (1/2) ln |1 + x2| + C {Using formula ∫f'(x)/f(x) dx = ln |f(x)| + C}
Hence, we have proved that the integral of tan inverse x is x tan-1x - (1/2) ln |1 + x2| + C, where C is the constant of integration.
Definite Integral of Tan Inverse x From 0 to 1
Now, we will determine the definite integral of arctan with limits from 0 to 1. We will use the integral of tan inverse x formula, that is, ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C to determine the definite integral.
\(\begin{align}\int_{0}^{1}\tan^{-1}x &=\left [ x\tan^{-1}x-\dfrac{1}{2}\ln|1+x^2|+C \right ]_0^1\\&=1\tan^{-1}1-\dfrac{1}{2}\ln|1+1^2|+C - (0\tan^{-1}0-\dfrac{1}{2}\ln|1+0^2|+C)\\&=\dfrac{\pi}{4}-\ln2+C-0+0-C\\&=\dfrac{\pi}{4}-\ln2 \end{align}\)
Hence, the definite integral of arctan from 0 to 1 is π/4 - ln 2.
Important Notes on Integral of Arctan
- ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C
- The definite integral of arctan from 0 to 1 is π/4 - ln 2
- Integral of arctan (x/a) is given by ∫tan-1(x/a) dx = x tan-1(x/a) - (a2/2) ln |a2 + x2| + C
Related Topics on Integral of Arctan
Integral of Arctan Examples
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Example 1: Find the integral of tan inverse root x.
Solution: To determine the integral of arctan root x, that is tan inverse root x, we will use integration by parts. We will use the following formulas:
Integration by parts - ∫f(x)g(x)dx = f(x) ∫g(x)dx - ∫[d(f(x))/dx × ∫g(x) dx] dx
d(tan-1√x)/dx = (1/2√x)[1/(1+x)]
∫(1/(1 + x2)) dx = tan-1x + C
∫tan-1√x dx = ∫1.tan-1√x dx
= tan-1√x ∫1dx - ∫[d(tan-1√x)/dx × ∫1 dx] dx
= x tan-1√x - ∫ {(1/2√x)[1/(1+x)]} x dx
= x tan-1√x - (1/2) ∫√x/(1+x) dx --- (1)
To simplify this, we will evaluate ∫√x/(1+x) dx using the substitution method.
Let √x = y ⇒ x = y2
⇒ dx = 2y dy, Substitute these values in ∫√x/(1+x) dx
∫√x/(1+x) dx = ∫y/(1+y2) 2y dy
= 2 ∫y2/(1 + y2) dy
= 2 ∫(y2 - 1 + 1)/(1 + y2) dy
= 2 ∫1 - (1/(1 + y2)) dy
= 2 [y - tan-1y] + C
= 2 [√x - tan-1√x] + C --- (2)
Substitute the value of ∫√x/(1+x) dx from (2) in (1)
∫tan-1√x dx = x tan-1√x - (1/2) ∫√x/(1+x) dx
= x tan-1√x - (1/2) × 2 [√x - tan-1√x] + C [From (2)]
= x tan-1√x - √x + tan-1√x + C
= (x + 1)tan-1√x - √x + C
Answer: The integral of tan inverse root x is (x + 1)tan-1√x - √x + C
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Example 2: Determine the integral of arctan (1/x).
Solution: We will find the integral of arctan (1/x) using the substitution method followed by integration by parts.
Let arctan (1/x) = θ ⇒ tan θ = 1/x ⇒ cot θ = x ⇒ -csc2θ = dx
Substitute the above values in ∫arctan (1/x) dx
∫arctan (1/x) dx = ∫θ(-csc2θ) dθ
= θ ∫-csc2θ dθ - ∫(d(θ)/dθ × ∫-csc2θ dθ) dθ
= θ cotθ - ∫cotθ dθ
= θ cotθ - ln |sinθ| + C
= x arctan (1/x) - ln|1/√(x2 + 1)| + C [Using tan θ = 1/x ⇒ sinθ = 1/√(x2 + 1)]
Answer: The integral of arctan (1/x) is x arctan (1/x) - ln|1/√(x2 + 1)| + C.
FAQs on Integral of Arctan
What is Integral of Arctan in Calculus?
The integral of arctan is the integration of tan inverse x which is given by ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C, where C is the constant of integration.
What is the Formula for Integral of Arctan?
The formula for integral of arctan is given by ∫tan-1x dx = x tan-1x - ½ ln |1+x2| + C.
How to Find the Integral of Arctan?
The integral of arctan can be calculated using the ILATE method, that is, integration by parts.
What is the Integral of Tan Inverse x From 0 to 1?
The definite integral of arctan from 0 to 1 is π/4 - ln 2.
Is the Integral of Arctan the Same as the Integral of Tan Inverse x?
Arctan is also called tan inverse. Therefore, the integral of arctan is the same as the integral of the tan inverse.
Is the Integral of Arctan equal to the Integral of Arccot?
Now, the integral of arctan is not equal to the integral of arccot as the integral of arctan is x tan-1x - ½ ln |1+x2| + C and the integral of arccot is x cot-1x + ½ ln |1+x2| + C.
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