Question

Solve $\int{\sec \left( \log x \right)}\left[ 1+\tan \left( \log x \right) \right]dx$
(a). $x\sec \left( \log x \right)+c$
(b). $\dfrac{x}{2}\sec \left( \log x \right)+c$
(c). $-x\sec \left( \log x \right)+c$
(d). $\dfrac{-x}{2}\sec \left( \log x \right)+c$

Answer 1

Hint: Suppose $\log x=t$ and take the base of the given $\log$ function as e,
So, we get
${{\log }_{e}}x=t$
So, ${{e}^{t}}=x$
Now, get the integral in terms of variable ‘t'. Use the following formula to get the answer.
$\int{{{e}^{x}}}\left[ f\left( x \right)+{{f}^{1}}\left( x \right) \right]dx={{e}^{x}}f\left( x \right)+c$
Where f(x) is any function.

Complete step-by-step answer:
Let us suppose the value of given integral is I. so, we get equation as
$I=\int{\sec \left( \log x \right)}\left[ 1+\tan \left( \log x \right) \right]dx$ …………. (i)
Suppose the value of $\log x$ is t. so, we get
$\log x=t$ ………… (ii)
As we know, ${{a}^{x}}=N$, then ${{\log }_{a}}N=x$. As by default, base of $\log x$ in equation is ‘e’. so, by the above mentioned rule $\left( {{a}^{x}}=N,\to {{\log }_{a}}N=x \right)$, we can write the equation (ii) as
${{\log }_{e}}x=t$
$x={{e}^{t}}$………….. (iii)
Now, differentiate the above expression, w.r.t ‘x’, we get
$\dfrac{d}{dx}x=\dfrac{d}{dx}\left( {{e}^{t}} \right)$
We know $\begin{aligned} & \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\dfrac{d}{dx}{{e}^{x}}={{e}^{x}} \\\ & 1={{e}^{t}}\dfrac{dt}{dx} \\\ \end{aligned}$
On cross-multiplying the above equation, we get
$dx={{e}^{t}}dt$
So, we can get equation (i) as
$\begin{aligned} & I=\int{\sec \left( t \right)}\left[ 1+\tan t \right]{{e}^{t}}dt \\\ & I=\int{{{e}^{t}}}\sec t\left[ 1+\tan t \right]dt \\\ \end{aligned}$
$I=\int{{{e}^{t}}\left[ \sec t+\tan t\sec t \right]}dt$………….. (iv)
Now, we know the integration of functions of type
$\int{{{e}^{x}}}\left( f\left( x \right)+{{f}^{1}}\left( x \right) \right)dx={{e}^{x}}f\left( x \right)$ …………… (v)
Now, compare the equation (iv) and right hand side of the equation (v). we get that $\sec t\tan t$ is derivative of $\sec t$, as we know
$\dfrac{d}{dx}\sec x=\sec x\tan x$ ………………(vi)
On comparing equation (iv) and (v), we get
$\begin{aligned} & I=\int{{{e}^{t}}}\left[ \sec t+\tan t\sec t \right]dt \\\ & I={{{e}^{t}}}\sec t+c \\\ \end{aligned}$
Now, we can put the value of ‘t’ from the equation (ii) and the value of ${{e}^{t}}$ from the equation (iii). So, we get value of I as
$I=x\sec \left( \log x \right)+c$
Hence, we get value of the given integral as
$\int{\sec \left( \log x \right)}\left[ 1+\tan \left( \log x \right) \right]dx=x\sec \left( \log x \right)+c$

Note: One may prove the identity of $\int{{{e}^{x}}}\left[ f\left( x \right)+{{f}^{1}}\left( x \right) \right]dx={{e}^{x}}f\left( x \right)$as
We have
$I=\int{{{e}^{x}}}\left( f\left( x \right)+{{f}^{1}}\left( x \right) \right)dx$
$I=\int{{{e}^{x}}}f\left( x \right)dx+\int{{{e}^{x}}}{{f}^{1}}\left( x \right)dx$
We integration by parts, which is given as
$\int{f\left( x \right)}g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)}dx-\int{{{f}^{1}}}\left( x \right)\int{g\left( x \right)dxdx}$
So, solve $\int{{{e}^{x}}f\left( x \right)}dx$ in the equation of I by integration by parts as
$\begin{aligned} & I=\int{{{e}^{x}}f\left( x \right)}dx+\int{{{e}^{x}}{{f}^{1}}\left( x \right)}dx \\\ & I=f\left( x \right)\int{{{e}^{x}}}dx-\int{{{f}^{1}}\left( x \right)\int{{{e}^{x}}}dx}dx+\int{{{e}^{x}}}{{f}^{1}}\left( x \right)dx \\\ & I=f\left( x \right){{e}^{x}}-\int{{{f}^{1}}\left( x \right){{e}^{x}}}dx+\int{{{e}^{x}}}{{f}^{1}}\left( x \right)dx \\\ & I={{e}^{x}}f\left( x \right) \\\ \end{aligned}$
So, one may use this identity with these kinds of questions directly.