How do you integrate $ {{\tan }^{3}}x{{\sec }^{5}}xdx $ ?

Hint: In this question, we need to find the integration of a function $ {{\tan }^{3}}x{{\sec }^{5}}x $ . For this we will first convert this into the terms of sinx and cosx. After that, we will rearrange the terms accordingly. As we can substitute cosx as u and find the integration easily. We will use the following formulas to solve this sum,
(I) tan is equal to $ \dfrac{\sin x}{\cos x} $ .
(II) secx is equal to $ \dfrac{1}{\cos x} $ .
(III) $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ .
(IV) $ \dfrac{d}{dx}\cos x=-\sin x $ .
(V) $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c} $ where c is any constant.

Complete step by step answer:
Here we are given the function as $ {{\tan }^{3}}x{{\sec }^{5}}x $ . We need to find the integration of this function i.e. we need to calculate $ \int{{{\tan }^{3}}x{{\sec }^{5}}xdx} $ .
For simplification let us convert the function in terms of sinx and cosx. We know that $ \tan x=\dfrac{\sin x}{\cos x}\text{ and }\sec x=\dfrac{1}{\cos x} $ . So the integral becomes $ \int{{{\left( \dfrac{\sin x}{\cos x} \right)}^{3}}}{{\left( \dfrac{1}{\cos x} \right)}^{5}}dx $ .
Simplifying it we get $ \int{\dfrac{{{\sin }^{3}}x}{{{\cos }^{8}}x}dx} $ .
Now we can write sinx as $ \sin x\cdot {{\sin }^{2}}x $ so we get $ \int{\dfrac{\sin x\cdot {{\sin }^{2}}x}{{{\cos }^{8}}x}}dx $ .
We know that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ . Rearranging it we get $ {{\sin }^{2}}x=1-{{\cos }^{2}}x $ . So the integral becomes $ \int{\dfrac{\sin x\cdot \left( 1-{{\cos }^{2}}x \right)}{{{\cos }^{8}}x}}dx $ .
To solve it easily we can substitute cosx as u. We have cosx = u.
Differentiating both sides we get $ -\sin xdx=du $ .
Taking negative to other side we get $ \sin xdx=-du $ .
Substituting the values in integral we get $ \int{\dfrac{-\left( 1-{{u}^{2}} \right)}{{{u}^{8}}}}du\Rightarrow \int{\dfrac{{{u}^{2}}-1}{{{u}^{8}}}}du $ .
Separating the numerator we get $ \int{\dfrac{{{u}^{2}}}{{{u}^{8}}}-\dfrac{1}{{{u}^{8}}}}du $ .
Separating the integrals we get $ \int{\dfrac{{{u}^{2}}}{{{u}^{8}}}du-\int{\dfrac{1}{{{u}^{8}}}}}du $ .
We know that \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\text{ and }{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] so we get $ \int{{{u}^{2-8}}du}-\int{{{u}^{-8}}du}\Rightarrow \int{{{u}^{-6}}du}-\int{{{u}^{-8}}du} $ .
Let us integrate them using $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c} $ we get $ \dfrac{{{u}^{-6+1}}}{-6+1}-\dfrac{{{u}^{-8+1}}}{-8+1}+c $ .
Simplifying the signs and rearranging the terms we get $ \dfrac{{{u}^{-5}}}{-5}-\dfrac{{{u}^{-7}}}{-7}+c $ .
Simplifying the denominator and power of u we get $ \dfrac{{{u}^{-7}}}{7}-\dfrac{{{u}^{-5}}}{5}+c $ .
Using \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] again we get $ \dfrac{1}{7{{u}^{7}}}-\dfrac{1}{5{{u}^{5}}}+c $ .
Now let us substitute value of u as cosx back we get $ \dfrac{1}{7{{\cos }^{7}}x}-\dfrac{1}{5{{\cos }^{5}}x}+c $ .
We know that $ \sec x=\dfrac{1}{\cos x} $ so we get $ \dfrac{{{\sec }^{7}}x}{7}-\dfrac{{{\sec }^{5}}x}{5}+c $ .
This is the required integration.
Hence $ \int{{{\tan }^{3}}x{{\sec }^{5}}xdx}=\dfrac{{{\sec }^{7}}x}{7}-\dfrac{{{\sec }^{5}}x}{5}+c $ .

Note:
Students should keep in mind all the trigonometric, integral properties to solve this sum. Take care of the signs for the power of u. Do not forget to substitute the value of u as cosx back. Do not forget to add the constant at the final answer.