Question

Evaluate the integrate $\int{\sin x\log \left( \sec x+\tan x \right)dx}=f\left( x \right)+x+c$, then find $f\left( x \right)$.
A. $\cos x\log \left( \sec x+\tan x \right)$
B. $\sin x\log \left( \sec x+\tan x \right)$
C. $-\cos x\log \left( \sec x+\tan x \right)$
D. $-\cos x\log \left( \sec x \right)$

Answer 1

We find the equation and put that in the by-parts theorem. We assign the functions as u and v of the theorem $\int{uvdx}=u\int{vdx}-\int{\dfrac{du}{dx}\left( \int{vdx} \right)dx}$. We separately find the differentiation value of $u=h\left( x \right)=\log \left( \sec x+\tan x \right)$. Then we put the values in the integral and find the actual integral solution. We equate the solution with the given term to find the value of the $f\left( x \right)$.

Complete step-by-step solution:
We have been given to find the integral value of $\sin x\log \left( \sec x+\tan x \right)$.
We are going to use the by parts theorem to find the integral value.
The theorem tells us for two functions $u=h\left( x \right),v=g\left( x \right)$.
$\int{uvdx}=u\int{vdx}-\int{\dfrac{du}{dx}\left( \int{vdx} \right)dx}$.
For our given integral we take $u=h\left( x \right)=\log \left( \sec x+\tan x \right),v=g\left( x \right)=\sin x$.
We try to find the differential form of the $u=h\left( x \right)=\log \left( \sec x+\tan x \right)$ for $\dfrac{du}{dx}$.
We have the differentiation of $\log x$ as $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$.
We also have $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x,\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$.
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( \log \left( \sec x+\tan x \right) \right)=\dfrac{\dfrac{d}{dx}\left( \sec x+\tan x \right)}{\sec x+\tan x}$
Now we do the rest of the differentiation
$\begin{aligned} & \Rightarrow \dfrac{\dfrac{d}{dx}\left( \sec x+\tan x \right)}{\sec x+\tan x} \\\ & \Rightarrow \dfrac{\sec x\tan x+{{\sec }^{2}}x}{\sec x+\tan x} \\\ & \Rightarrow \dfrac{\sec x\left( \sec x+\tan x \right)}{\left( \sec x+\tan x \right)} \\\ & \Rightarrow \sec x \\\ \end{aligned}$
So, in the by-parts we will directly use the value of $\dfrac{du}{dx}=\dfrac{d}{dx}\left( \log \left( \sec x+\tan x \right) \right)=\sec x$.
We also have $\int{\left( \sin x \right)dx}=-\cos x$.
$\begin{aligned} & \int{\sin x\log \left( \sec x+\tan x \right)dx} \\\ & \Rightarrow \left[ \log \left( \sec x+\tan x \right) \right]\int{\left( \sin x \right)dx}-\int{\sec x\left( \int{\left( \sin x \right)dx} \right)dx} \\\ & \Rightarrow -\cos x\left[ \log \left( \sec x+\tan x \right) \right]+\int{\sec x.\cos xdx} \\\ & \Rightarrow -\cos x\left[ \log \left( \sec x+\tan x \right) \right]+\int{dx} \\\ & \Rightarrow -\cos x\left[ \log \left( \sec x+\tan x \right) \right]+x+c \\\ \end{aligned}$
c is the integral constant.
Now we have to equate the given function on the right-hand side of $\int{\sin x\log \left( \sec x+\tan x \right)dx}=f\left( x \right)+x+c$ with the solution.
So, $f\left( x \right)+x+c=-\cos x\left[ \log \left( \sec x+\tan x \right) \right]+x+c$.
We need to find the value of $f\left( x \right)$. Equating the both sides we get $f\left( x \right)=-\cos x\left[ \log \left( \sec x+\tan x \right) \right]$.
The correct option is C.

Note: We need to be careful about choosing the functions carefully. If we would have taken
$v=g\left( x \right)=\log \left( \sec x+\tan x \right)$, finding the integral of the function would have been difficult. Also, before starting the solution we need to work out these differentiation and small integrals to find out in which direction the solution will move.