Integrate the function \[{\tan ^2}\left( {2x - 3} \right)\]
Answer
620.7k+ 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
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

