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Evaluate $2\int {{{\tan }^4}t{{\sec }^2}tdt} $.

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Last updated date: 22nd May 2024
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Answer
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Hint: The given question requires us to integrate a trigonometric function of $t$ with respect to $t$. Integration gives us a family of curves. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc. integral is always found with respect to some variables, which in this case is $x$.

Complete step by step solution:
The given question requires us to integrate a trigonometric function ${\tan ^4}t{\sec ^2}t$ in variable t.
Now, the integral given to us is complex and cannot be solved directly using basic results of integration of functions. We will have to simplify the function so as to solve the integral.
So, we can assign a new variable in the integral.
Let us assume $\tan t = x$.
Then, differentiating both sides of the equation, we get,
$ \Rightarrow {\sec ^2}tdt = dx$
Hence, the integral given is,
$2\int {{{\tan }^4}t{{\sec }^2}tdt} $
Substituting the value of ${\sec ^2}tdt$ in terms of x as ${\sec ^2}tdt = dx$ and the value of $\tan t$ as x, we get,
$ \Rightarrow 2\int {{x^4}dx} $
Now, we know that the integral of ${x^n}$ with respect to x is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$ using the power rule of integration. So, we get,
$ \Rightarrow 2\left( {\dfrac{{{x^5}}}{5}} \right) + c$, where c is an arbitrary constant.
Substituting the value of x in the expression as $\tan t$, we get,
$ \Rightarrow 2\left( {\dfrac{{{{\tan }^5}t}}{5}} \right) + c$
Opening the bracket, we get,
$ \Rightarrow \dfrac{{2{{\tan }^5}t}}{5} + c$
So, $\dfrac{{2{{\tan }^5}t}}{5} + c$, where c is any arbitrary constant, is the value of the given integral, $2\int {{{\tan }^4}t{{\sec }^2}tdt} $.

Note:
The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant.