Question

How do you find $\int {x \cdot \sec {x^2}dx}$?

Answer 1

We can use Integration by substitution to solve this integral. So, first substitute $u = {x^2}$ and find its differentiation with respect to $x$ using differentiation properties. Then substitute the value of $x$ and value of $dx$ in the given integral. Next, integrate using integration properties. Finally, substitute the value of $u$, and get the desired result.

Formula used:
The differentiation of the product of a constant and a function = the constant $\times$ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
Differentiation formula: $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$
The integral of the product of a constant and a function = the constant $\times$ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx}$, where $k$ is a constant.
The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx}$
Integration formula: $\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right| + C$

Complete step by step answer:
We have to find $\int {x \cdot \sec {x^2}dx}$…(i)
We will use Integration by substitution to solve this integral.
So, for the integral involving the square${x^2}$, we use the substitution:
$u = {x^2}$…(ii)
Now, we have to differentiate $u$ with respect to $x$.
So, differentiating $u$ with respect to $x$, we get
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}} \right)$…(iii)
Now, using the differentiation formula $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$ in above differentiation and find the value of $dx$.
$\Rightarrow \dfrac{{du}}{{dx}} = 2x$
$\Rightarrow dx = \dfrac{1}{{2x}}du$…(iv)
Now, we have to substitute the value of $\sqrt x$ from (ii) and the value of $dx$ from (iv) in integral (i).
$\int {x \cdot \sec {x^2}dx} = \int {x \cdot \sec u \times \dfrac{1}{{2x}}du}$
$\Rightarrow \int {x \cdot \sec {x^2}dx} = \int {\dfrac{1}{2}\sec udu}$…(v)
Now, using the property that the integral of the product of a constant and a function = the constant $\times$ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx}$, where $k$ is a constant.
So, in above integral (v), constant $2$ can be taken outside the integral.
$\Rightarrow \int {x \cdot \sec {x^2}dx} = \dfrac{1}{2}\int {\sec udu}$…(vi)
Now, using the integration formula $\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right| + C$ in integral (vi), we get
$\Rightarrow \int {x \cdot \sec {x^2}dx} = \dfrac{1}{2}\ln \left| {\sec u + \tan u} \right| + C$…(vii)
We have to find the integral in terms of $x$. So, replacing $u$ with $x$ using (ii).
$\Rightarrow \int {x \cdot \sec {x^2}dx} = \dfrac{1}{2}\ln \left| {\sec {x^2} + \tan {x^2}} \right| + C$

Hence, $\int {x \cdot \sec {x^2}dx} = \dfrac{1}{2}\ln \left| {\sec {x^2} + \tan {x^2}} \right| + C$.

Note: Integration by Substitution:
Integrals of certain functions cannot be obtained directly if they are not in one of the standard forms, but they may be reduced to standard forms by proper substitution. The method of evaluating an integral by reducing it to standard form by a proper substitution is called integration by substitution.
If $\phi \left( x \right)$ is a continuously differentiable function, then to evaluate integrals of the form
$\int {f\left( {\phi \left( x \right)} \right)\phi '\left( x \right)dx}$,
We substitute $\phi \left( x \right) = t$ and $\phi '\left( x \right)dx = dt$.
These substitutions reduce the above integral to $\int {f\left( t \right)dt}$. After evaluating this integral we substitute back the value of $t$.