Answer
Verified
429.3k+ views
Hint:We are asked to integrate the given function. The angle of tangent is given as \[\left( {2x - 3} \right)\] so for simplification assume the term \[\left( {2x - 3} \right)\] as a whole to be a number. Then use the trigonometric identities to simplify the problem and then use the basic formulas for integration to solve the problem.
Complete step by step solution:
Given the function \[{\tan ^2}\left( {2x - 3} \right)\]
Now we integrate the function,
\[I = \int {{{\tan }^2}\left( {2x - 3} \right)dx} \] (i)
Let \[2x - 3 = t\] (ii)
Differentiating equation (ii), we get
\[2dx = dt\]
\[ \Rightarrow dx = \dfrac{1}{2}dt\] (iii)
Using equation (iii) and (ii) in (i) we get,
\[I = \int {\left( {{{\tan }^2}t} \right)\left( {\dfrac{1}{2}dt} \right)} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {{{\tan }^2}tdt} \] (iv)
We have the trigonometric identity for tangent as,
\[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
\[ \Rightarrow {\tan ^2}\theta = {\sec ^2}\theta - 1\]
Therefore using this formula for \[{\tan ^2}t\] we have,
\[{\tan ^2}t = {\sec ^2}t - 1\]
Substituting this value of \[{\tan ^2}t\] in equation (iv) we get,
\[I = \dfrac{1}{2}\int {\left( {{{\sec }^2}t - 1} \right)dt} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {{{\sec }^2}tdt} - \dfrac{1}{2}\int {dt} \] (v)
Integration of \[{\sec ^2}\theta \] is \[\tan \theta \]. Therefore using this in equation (v) we get,
\[I = \dfrac{1}{2}\tan t - \dfrac{1}{2}t + c\] (vi)
\[c\] is constant of integration
Now, substituting equation (ii) in (vi), we get
\[I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - \dfrac{1}{2}\left( {2x - 3} \right) + c\]
\[ \Rightarrow I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + \dfrac{3}{2} + c\] (vii)
The term \[\left( {\dfrac{3}{2} + c} \right)\] can be considered as constant so, we can write equation (vii) as,
\[I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + C\]
where \[C\] is constant.
Therefore, integration of the function \[{\tan ^2}\left( {2x - 3} \right)\] is \[\dfrac{1}{2}\tan \left( {2x - 3}\right) - x + C\].
Note: Integration means adding small components to form a whole function. Here, we were asked to integrate a trigonometric function, to solve such types of questions, you will need to remember the integration of basic trigonometric functions that are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the basic functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic trigonometric identities.
Complete step by step solution:
Given the function \[{\tan ^2}\left( {2x - 3} \right)\]
Now we integrate the function,
\[I = \int {{{\tan }^2}\left( {2x - 3} \right)dx} \] (i)
Let \[2x - 3 = t\] (ii)
Differentiating equation (ii), we get
\[2dx = dt\]
\[ \Rightarrow dx = \dfrac{1}{2}dt\] (iii)
Using equation (iii) and (ii) in (i) we get,
\[I = \int {\left( {{{\tan }^2}t} \right)\left( {\dfrac{1}{2}dt} \right)} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {{{\tan }^2}tdt} \] (iv)
We have the trigonometric identity for tangent as,
\[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
\[ \Rightarrow {\tan ^2}\theta = {\sec ^2}\theta - 1\]
Therefore using this formula for \[{\tan ^2}t\] we have,
\[{\tan ^2}t = {\sec ^2}t - 1\]
Substituting this value of \[{\tan ^2}t\] in equation (iv) we get,
\[I = \dfrac{1}{2}\int {\left( {{{\sec }^2}t - 1} \right)dt} \]
\[ \Rightarrow I = \dfrac{1}{2}\int {{{\sec }^2}tdt} - \dfrac{1}{2}\int {dt} \] (v)
Integration of \[{\sec ^2}\theta \] is \[\tan \theta \]. Therefore using this in equation (v) we get,
\[I = \dfrac{1}{2}\tan t - \dfrac{1}{2}t + c\] (vi)
\[c\] is constant of integration
Now, substituting equation (ii) in (vi), we get
\[I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - \dfrac{1}{2}\left( {2x - 3} \right) + c\]
\[ \Rightarrow I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + \dfrac{3}{2} + c\] (vii)
The term \[\left( {\dfrac{3}{2} + c} \right)\] can be considered as constant so, we can write equation (vii) as,
\[I = \dfrac{1}{2}\tan \left( {2x - 3} \right) - x + C\]
where \[C\] is constant.
Therefore, integration of the function \[{\tan ^2}\left( {2x - 3} \right)\] is \[\dfrac{1}{2}\tan \left( {2x - 3}\right) - x + C\].
Note: Integration means adding small components to form a whole function. Here, we were asked to integrate a trigonometric function, to solve such types of questions, you will need to remember the integration of basic trigonometric functions that are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the basic functions, these are cosecant which is inverse of sine, secant which is inverse of cosine and cotangent which is inverse of tangent. Also, while solving questions related to trigonometry, you should always remember the basic trigonometric identities.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
If the mean of the set of numbers x1x2xn is bar x then class 10 maths JEE_Main
What is the meaning of celestial class 10 social science CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE