Question

How do you integrate $\int {\;{{\sec }^4}\left( {5x} \right)dx?}$

Answer 1

In this question we have simplified the given integration. Next, we use some trigonometric identities and then simplify to arrive at our final answer. Next, we integrate the trigonometry functions .And also we are going to multiplication and addition in complete step by step solution.
Calculus, branch of mathematics concerned with the calculation of instantaneous rates of change (differential calculus) and the summation of infinitely many small factors to determine some whole (integral calculus).
The trigonometric functions (also called circular functions, angle functions or goniometric functions) are real functions which relate an angle of a right-angled triangle to ratios of two side lengths.

Complete step-by-step solution:
Given,
$\Rightarrow \int {\;{{\sec }^4}\left( {5x} \right)dx}$
Start first with a simple substitution to simplify the argument of the secant functions: let $u = 5x$ which implies that$du = 5dx$. Then:
$\Rightarrow \dfrac{1}{5}\int {{{\sec }^4}(5x)(5dx)}$
$\Rightarrow \dfrac{1}{5}\int {{{\sec }^4}(u)du}$
When working with secant, it's important to keep the identities $\dfrac{d}{{dx}}\sec x = \sec x\tan x$,$\dfrac{d}{{dx}}\tan x = {\sec ^2}x$ and $1 + {\tan ^2}x = {\sec ^2}x$in mind.
In this case, we see that the integral
$\Rightarrow \dfrac{1}{5}\int {{{\sec }^2}(u){{\sec }^2}(u)du}$
Put, $1 + {\tan ^2}x = {\sec ^2}x$ in the above term and we get
$\Rightarrow \dfrac{1}{5}\int {\left( {1 + {{\tan }^2}\left( u \right)} \right){{\sec }^2}(u)du}$
The point of doing this is that we have a function of tangent$1 + {\tan ^2}x$paired with the derivative of tangent ${\sec ^2}x$.
So, let $v = \tan (u)$ so that $dv = {\sec ^2}(u)du$. This gives us:
$\Rightarrow \dfrac{1}{5}\int {\left( {1 + {v^2}} \right)dv}$
Next, split the integral term in the above term and we get
$\Rightarrow \dfrac{1}{5}\left( {\int {1dv + \int {{v^2}} dv} } \right)$
We can now integrate term by term:
$\Rightarrow \dfrac{1}{5}\left( {v + \dfrac{{{v^3}}}{3}} \right) + c$
Now, multiply 1/5 into the bracket and we get
$\Rightarrow \left( {\dfrac{v}{5} + \dfrac{{{v^3}}}{{3 \times 5}}} \right) + c$
On multiply the term and we get,
$\Rightarrow \left( {\dfrac{v}{5} + \dfrac{{{v^3}}}{{15}}} \right) + c$
Returning to our original variable $x$ using $v = \tan (u)$ and $u = 5x$ then adding the constant of integration:
$\Rightarrow \left( {\dfrac{{\tan u}}{5} + \dfrac{{{{\tan }^3}u}}{{15}}} \right) + c$
$\Rightarrow \left( {\dfrac{{\tan \left( {5x} \right)}}{5} + \dfrac{{{{\tan }^3}\left( {5x} \right)}}{{15}}} \right) + c$ Here, $u = 5x$

$\left( {\dfrac{{\tan \left( {5x} \right)}}{5} + \dfrac{{{{\tan }^3}\left( {5x} \right)}}{{15}}} \right) + c$ is the required integrate value of $\int {\;{{\sec }^4}\left( {5x} \right)dx}$

Note: We have to mind that, Integration, in mathematics, technique of finding a function $g\left( x \right)$ the derivative of which,$Dg\left( x \right)$ is equal to a given function$f\left( x \right)$. This is indicated by the integral sign, as in $\smallint f\left( x \right)$, usually called the indefinite integral of the function. The symbol $dx$ represents an infinitesimal displacement along $x$; thus $\smallint f\left( x \right)dx$ is the summation of the product of $f\left( x \right)$ and $dx$.
The definite integral, written $\int\limits_a^b {f\left( x \right)dx}$ with $a$ and $b$ called the limits of integration is equal to $g\left( b \right){\text{ }} - g\left( a \right)$, where $Dg\left( x \right){\text{ }} = f\left( x \right)$.
If you're going into fields of science such as physics, chemistry, engineering, or higher mathematics, calculus is crucial. Calculus is the study of rates of change of things that algebra alone can't fully explain. Calculus is also linked very strongly to areas and volumes of shapes and solids.
Calculus is mostly important in the sciences, but if you look around you, you can see other applications of calculus inside and outside your home.