Solveeit Logo
Login

Question

Question: How do you integrate \( \int {\arctan \left( {\dfrac{1}{x}} \right)} \) using integration by parts?...

How do you integrate ∫arctan⁡(1x)\int {\arctan \left( {\dfrac{1}{x}} \right)} using integration by parts?

Explanation

Solution

Hint : In mathematics, integration is the concept of calculus and it is the act of finding the integrals. Here we will find integration, by using the concept of equivalent value, and will simplify and then will place the formula and simplify for the resultant answer. Here we will assume first for the part of integration, and then find its derivative and the function of the given integral. Here we will use any variable as the reference in the above expression, will place it in the above equation and will frame the simplified equation and will convert the given expression in the form of “u” into “v” and then will find integration by solving in parts.

Complete step by step solution:
Take the given expression:
I=∫arctan⁡(1x)dxI = \int {\arctan \left( {\dfrac{1}{x}} \right)} dx …. (A)
Let us assume that θ=arctan⁡(1x)\theta = \arctan \left( {\dfrac{1}{x}} \right) …. (B)
Cross multiplication implies –
tan⁡θ=(1x)\tan \theta = \left( {\dfrac{1}{x}} \right)
We know that tangent and cot are the inverse functions of each other.
⇒cot⁡θ=x\Rightarrow \cot \theta = x
Differentiate above equation with respect to θ\theta
⇒ddθcot⁡θ=dxdθ\Rightarrow \dfrac{d}{{d\theta }}\cot \theta = \dfrac{{dx}}{{d\theta }}
Place the values of the differentiation in the above equation.
⇒−cos⁡ec2θ=dxdθ\Rightarrow - \cos e{c^2}\theta = \dfrac{{dx}}{{d\theta }}
Do cross multiplication, where the denominator of one side is multiplied with the numerator on the opposite side.
⇒−cos⁡ec2θdθ=dx\Rightarrow - \cos e{c^2}\theta d\theta = dx …. (c)
Place values of the equation (B) and (C) in the equation (A)
I=∫θ(−cos⁡ec2θ)dθI = \int {\theta ( - \cos e{c^2}\theta )d\theta }
Here we will use, the integration by parts formula
Let us assume that, u=θu = \theta and dv=−cos⁡ec2θdθdv = - \cos e{c^2}\theta d\theta
So, du=dθdu = d\theta and v=cot⁡θv = \cot \theta
Using above relations and correlations,
uv−∫vdu=θcot⁡θ−∫cot⁡θdθuv - \int {vdu = \theta \cot \theta - \int {\cot \theta d\theta } }
Substitute the integration formula,
uv−∫vdu=θcot⁡θ−ln⁡∣sin⁡θ∣+Cuv - \int {vdu = \theta \cot \theta - \ln \left| {\sin \theta } \right|} + C
Now, using cot⁡θ=x\cot \theta = x
sin⁡θ=1x2+1\sin \theta = \dfrac{1}{{\sqrt {{x^2} + 1} }}
Replacing the actual values in the equation (A)
∫arctan⁡(1x)dx=xarctan⁡(1x)−ln⁡(1x2+1)+c\int {\arctan \left( {\dfrac{1}{x}} \right)} dx = x\arctan \left( {\dfrac{1}{x}} \right) - \ln \left( {\dfrac{1}{{\sqrt {{x^2} + 1} }}} \right) + c
This is the required solution.
So, the correct answer is “ xarctan⁡(1x)−ln⁡(1x2+1)+cx\arctan \left( {\dfrac{1}{x}} \right) - \ln \left( {\dfrac{1}{{\sqrt {{x^2} + 1} }}} \right) + c ”.

Note : Know the difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverse of each other.