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Question: What is the integral of \[\dfrac{1}{\tan\left( x \right)}{\ dx}\] ?...

What is the integral of 1tan(x) dx\dfrac{1}{\tan\left( x \right)}{\ dx} ?

Explanation

Solution

In this question, we need to find the integral of 1tan(x) dx\dfrac{1}{\tan\left( x \right)}{\ dx}. Sine , cosine and tangent are the basic trigonometric functions . Here with the help of the basic trigonometric identities and ratios , we can solve this question . Tangent function is nothing but it is defined as a ratio of the opposite side of a right angle to the adjacent side of the right angle. The symbol '\int’ is known as the sign of integration. The process of finding the integral is known as integration. The methods used to integrate the given expression is reciprocal rule and reverse chain rule. This method is related to the chain rule of differentiation, which when applied to antiderivatives is known as the reverse chain rule that is integration by u substitution.
Formula used :
cotθ=cosθsinθ\cot\theta = \dfrac{\cos\theta}{sin\theta}
1tanθ=cotθ\dfrac{1}{tan\theta} = cot\theta
Reciprocal rule :
(1x)dx=lnx+ c\int\left( \dfrac{1}{x} \right)dx = ln\left| x \right| + \ c
Where cc is the constant of integration.
Derivative rule used :
d(sinx)dx=cos x\dfrac{d\left( {sinx} \right)}{{dx}} = cos\ x
Reverse chain rule :
Reverse chain rule is also known as u-Substitution. U sub is a special method of integration. It is applicable, when the expression contains two functions. This method combines two functions with the help of another variable u‘u’ and makes the integration process direct and much easier.
 f(x) f(x) dx\int\ f(x)\ f’(x)\ dx
The original component f(x)f(x)
The derivative component f(x) dxf’(x)\ dx
 f(x) f(x) dx=u du\int\ f(x)\ f’(x)\ dx = \int u\ du

Complete step-by-step solution:
Given,
1tan(x) dx\dfrac{1}{\tan\left( x \right)}{\ dx}
Here need to find the integral of 1tan(x) dx\dfrac{1}{\tan\left( x \right)}{\ dx}
Let us consider I=1tan(x) dxI = \dfrac{1}{\tan\left( x \right)}{\ dx}
We know that 1tanθ=cotθ\dfrac{1}{{tan\theta}} = cot\theta
Thus we get, I=cot x dxI = cot\ x\ {dx}
We also know that, cotθ=cosθsinθ\cot\theta = \dfrac{\cos\theta}{sin\theta}
Thus we get, I=cosxsinxdxI = \int\dfrac{{cosx}}{{sinx}}{dx} ••• (1)
Let us consider u=sin xu = sin\ x
On differentiating u=sin xu = sin\ x,
We know that d(sinx)dx=cos x\dfrac{d\left( {sinx} \right)}{{dx}} = cos\ x
We get, dudx=cosx\dfrac{{du}}{{dx}} = cosx
 du=cosx dx\Rightarrow\ du = cosx\ dx
Thus the equation (1) becomes,
I=1uduI = \int\dfrac{1}{u}{du}
We know that
(1x)dx=lnx+ c\int\left( \dfrac{1}{x} \right)dx = ln\left| x \right| + \ c
Therefore we get, I=lnu+c\ I = ln|u| + c
Where cc is the constant of integration.
By substituting the value of u=sin xu = sin\ x,
We get,
I=lnsinx+cI = ln|sinx| + c
Hence we get the integral of 1tan(x)dx \dfrac{1}{\tan\left( x \right){dx}}\ is lnsinx+cln|sinx| + c
Final answer :
The integral of 1tan(x)dx \dfrac{1}{\tan\left( x \right){dx}}\ is lnsinx+cln|sinx| + c

Note: The concept used in this question is integration method, that is integration by substitution and also with the help of reverse chain rule we can find the integration of the given expression . Since this is an indefinite integral we have to add an arbitrary constant `cc’. cc is called the constant of integration. The variable xx in dx{dx} is known as the variable of integration or integrator. In this question, the derivative rule is also used to solve. Mathematically, a derivative is defined as a rate of change of function with respect to an independent variable given in the function.