
What is the integral of \[\int \tan(2x)\ dx\] ?
Answer
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Hint: In this question, we need to find the integral of \[\int \tan(2x)\ dx\] . Integration is nothing but its derivative is equal to its original function. First we can consider the given expression as \[I\] . Then we can use the reverse chain rule by substituting the function as \[u\] . After integrating, then we need to do some rearrangements of terms and hence we can find the integral of the given expression.
Formula used :
1.\[\dfrac{d}{dx}(\text{cosx}) = - \sin\ x\]
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 ‘\[u\]’ and makes the integration process direct and much easier.
\[\int\ f(x)\ f’(x)\ dx\]
Where the original component is \[f(x)\]
The derivative component is \[f’(x)\ dx\]
\[\int\ f(x)\ f’(x)\ dx = \int u\ du\]
Complete step by step answer:
Given, \[\int \tan(2x)\ dx\]
Let us consider, \[I = \int \tan(2x)\ dx\]
We know that \[\tan(x)\ = \dfrac{\sin\left( x \right)}{\cos\left( x \right)}\]
Therefore, we can rewrite \[\tan(2x)\] as \[\dfrac{\sin\left( 2x \right)}{\cos\left( 2x \right)}\] ,
Thus we get,
\[I = \int \tan\left( 2x \right){dx} = \int\dfrac{\sin\left( 2x \right)}{\cos\left( 2x \right)}{dx}\]
On multiplying and dividing both the numerator and denominator by \[2\] ,
We get,
\[I = - \dfrac{1}{2}\int\left( \dfrac{- 2\sin\left( 2x \right)}{\cos\left( 2x \right)} \right){dx}\] ••• (1)
Let us consider, \[u = cos(2x)\]
On differentiating \[u\] ,
We get,
\(\dfrac{du}{dx} = - 2\sin(2x)\)
\[\Rightarrow \ du = - 2\sin(2x)\ dx\]
Thus the equation (1) becomes,
\[I = - \dfrac{1}{2}\ \int\left( \dfrac{{du}}{u} \right)\]
We know that \[\int\left( \dfrac{1}{x} \right)dx = ln\left| x \right| + \ c\] Where \[c\] is the constant of integration.
We get,
\[I = - \dfrac{1}{2}\left( \ln\left| u \right| + c \right)\]
By substituting the value of \[u\] ,
We get,
\[I = - \dfrac{1}{2}\ln\left| \cos\left( 2x \right) \right| + c\] where \[c\] is the constant of integration .
Thus we get the integral of \[\int \tan(2x)\ dx\] is \[- \dfrac{1}{2}\ln\left| \cos\left( 2x \right) \right| + \ c\]
The integral of \[\int \tan(2x)\ dx\] is \[- \dfrac{1}{2}\ln\left| \cos\left( 2x \right) \right| + \ c\]
Note: The concept used in this question is integration method, that is integration by u-substitution and reciprocal rule . Since this is an indefinite integral we have to add an arbitrary constant ‘\[c\]’. \[c\] is called the constant of integration. The variable \[x\] in \[dx\] is known as the variable of integration or integrator. Constant integration is very much important while writing the integral value, because the constant term always has derivative zero. The reverse chain rule 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. Mathematically, integrals are also used to find many useful quantities such as areas, volumes, displacement, etc.
Formula used :
1.\[\dfrac{d}{dx}(\text{cosx}) = - \sin\ x\]
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 ‘\[u\]’ and makes the integration process direct and much easier.
\[\int\ f(x)\ f’(x)\ dx\]
Where the original component is \[f(x)\]
The derivative component is \[f’(x)\ dx\]
\[\int\ f(x)\ f’(x)\ dx = \int u\ du\]
Complete step by step answer:
Given, \[\int \tan(2x)\ dx\]
Let us consider, \[I = \int \tan(2x)\ dx\]
We know that \[\tan(x)\ = \dfrac{\sin\left( x \right)}{\cos\left( x \right)}\]
Therefore, we can rewrite \[\tan(2x)\] as \[\dfrac{\sin\left( 2x \right)}{\cos\left( 2x \right)}\] ,
Thus we get,
\[I = \int \tan\left( 2x \right){dx} = \int\dfrac{\sin\left( 2x \right)}{\cos\left( 2x \right)}{dx}\]
On multiplying and dividing both the numerator and denominator by \[2\] ,
We get,
\[I = - \dfrac{1}{2}\int\left( \dfrac{- 2\sin\left( 2x \right)}{\cos\left( 2x \right)} \right){dx}\] ••• (1)
Let us consider, \[u = cos(2x)\]
On differentiating \[u\] ,
We get,
\(\dfrac{du}{dx} = - 2\sin(2x)\)
\[\Rightarrow \ du = - 2\sin(2x)\ dx\]
Thus the equation (1) becomes,
\[I = - \dfrac{1}{2}\ \int\left( \dfrac{{du}}{u} \right)\]
We know that \[\int\left( \dfrac{1}{x} \right)dx = ln\left| x \right| + \ c\] Where \[c\] is the constant of integration.
We get,
\[I = - \dfrac{1}{2}\left( \ln\left| u \right| + c \right)\]
By substituting the value of \[u\] ,
We get,
\[I = - \dfrac{1}{2}\ln\left| \cos\left( 2x \right) \right| + c\] where \[c\] is the constant of integration .
Thus we get the integral of \[\int \tan(2x)\ dx\] is \[- \dfrac{1}{2}\ln\left| \cos\left( 2x \right) \right| + \ c\]
The integral of \[\int \tan(2x)\ dx\] is \[- \dfrac{1}{2}\ln\left| \cos\left( 2x \right) \right| + \ c\]
Note: The concept used in this question is integration method, that is integration by u-substitution and reciprocal rule . Since this is an indefinite integral we have to add an arbitrary constant ‘\[c\]’. \[c\] is called the constant of integration. The variable \[x\] in \[dx\] is known as the variable of integration or integrator. Constant integration is very much important while writing the integral value, because the constant term always has derivative zero. The reverse chain rule 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. Mathematically, integrals are also used to find many useful quantities such as areas, volumes, displacement, etc.
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