
What is the integral of $\int{{{\tan }^{4}}\left( x \right){{\sec }^{2}}\left( x \right)dx}$?
Answer
516.3k+ views
Hint: Assume the given integral as ‘I’. Now, substitute $\tan x=k$ and differentiate both the sides to find the value of ${{\sec }^{2}}xdx$ in terms of $dk$ and substitute in the integral. Use the formula $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$. Use the basic formula of the integral given as $\int{{{k}^{n}}dk}=\dfrac{{{k}^{n+1}}}{n+1}$ to get the answer. Substitute back the assumed value of k to get the function in terms of x. Finally, add the constant of indefinite integration (c) at last.
Complete step by step answer:
Here we have been provided with the function ${{\tan }^{4}}\left( x \right){{\sec }^{2}}\left( x \right)$ and we are asked to integrate it. Let us assume the integral as I, so we have,
$\Rightarrow I=\int{{{\tan }^{4}}\left( x \right){{\sec }^{2}}\left( x \right)dx}$
Let us use the substitution method to solve this integral, so substituting $\tan x=k$ and differentiating both the sides to find the value of ${{\sec }^{2}}xdx$ in terms of $dk$ we get,
$\begin{align}
& \Rightarrow d\left( \tan x \right)=dk \\
& \Rightarrow {{\sec }^{2}}xdx=dk \\
\end{align}$
Substituting the assumed and above obtained relation in the integral I we get,
\[\Rightarrow I=\int{{{k}^{4}}dk}\]
Using the formula $\int{{{k}^{n}}dk}=\dfrac{{{k}^{n+1}}}{n+1}$ where n must not be equal to -1, we get,
\[\begin{align}
& \Rightarrow I=\dfrac{{{k}^{4+1}}}{4+1} \\
& \Rightarrow I=\dfrac{{{k}^{5}}}{5} \\
\end{align}\]
Substituting back the value of k we get,
\[\Rightarrow I=\dfrac{{{\tan }^{5}}x}{5}\]
Now, since the given integral is an indefinite integral and therefore we need to add a constant of integration (c) in the expression obtained for I. So we get,
\[\Rightarrow I=\dfrac{{{\tan }^{5}}x}{5}+c\]
Hence, the above relation is our answer.
Note: Note that it will be difficult to find the required integral without the help of a substitution method. This method is generally used when we get a hint while observing the integral that we have the function and its derivative term inside the integral sign. Remember the derivative formula of all the basic functions like: trigonometric and inverse trigonometric functions, exponential functions, log functions etc.
Complete step by step answer:
Here we have been provided with the function ${{\tan }^{4}}\left( x \right){{\sec }^{2}}\left( x \right)$ and we are asked to integrate it. Let us assume the integral as I, so we have,
$\Rightarrow I=\int{{{\tan }^{4}}\left( x \right){{\sec }^{2}}\left( x \right)dx}$
Let us use the substitution method to solve this integral, so substituting $\tan x=k$ and differentiating both the sides to find the value of ${{\sec }^{2}}xdx$ in terms of $dk$ we get,
$\begin{align}
& \Rightarrow d\left( \tan x \right)=dk \\
& \Rightarrow {{\sec }^{2}}xdx=dk \\
\end{align}$
Substituting the assumed and above obtained relation in the integral I we get,
\[\Rightarrow I=\int{{{k}^{4}}dk}\]
Using the formula $\int{{{k}^{n}}dk}=\dfrac{{{k}^{n+1}}}{n+1}$ where n must not be equal to -1, we get,
\[\begin{align}
& \Rightarrow I=\dfrac{{{k}^{4+1}}}{4+1} \\
& \Rightarrow I=\dfrac{{{k}^{5}}}{5} \\
\end{align}\]
Substituting back the value of k we get,
\[\Rightarrow I=\dfrac{{{\tan }^{5}}x}{5}\]
Now, since the given integral is an indefinite integral and therefore we need to add a constant of integration (c) in the expression obtained for I. So we get,
\[\Rightarrow I=\dfrac{{{\tan }^{5}}x}{5}+c\]
Hence, the above relation is our answer.
Note: Note that it will be difficult to find the required integral without the help of a substitution method. This method is generally used when we get a hint while observing the integral that we have the function and its derivative term inside the integral sign. Remember the derivative formula of all the basic functions like: trigonometric and inverse trigonometric functions, exponential functions, log functions etc.
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