Question

Integrate the given expression,
\[\int{\sec x.\log \left( \sec x+\tan x \right)dx}\]

Answer 1

Hint: Put the second term \[\log \left( \sec x+\tan x \right)=t\]. Thus differentiate and get the value of \[\dfrac{dt}{dx}\]. Thus the given expression changes in terms of t after substitution. Thus integrate the expression and replace the value of t.

Complete step-by-step answer:
We have been given the expression, which we need to find the integration. Let us put it as I.
\[I=\int{\sec x.\log \left( \sec x+\tan x \right)dx}-(1)\]
Let us consider, \[t=\log \left( \sec x+\tan x \right)-(2)\].
Thus let us differentiate and get the value of \[\dfrac{dt}{dx}\].
We know, \[\dfrac{d}{dx}\log x=\dfrac{1}{x}\].
Similarly, \[\log \left( \sec x+\tan x \right)=\dfrac{1}{\left( \sec x+\tan x \right)}\], then differentiate \[\left( \sec x+\tan x \right)\].
The differentiation of, \[\sec x=\sec x\tan x\].
Similarly, the differentiation of \[\tan x={{\sec }^{2}}x\].
\[\begin{align}
  & \therefore \dfrac{dt}{dx}=\dfrac{\sec x\tan x+{{\sec }^{2}}x}{\sec x+\tan x} \\
 & \therefore \dfrac{dt}{dx}=\dfrac{\sec x\left( \tan x+\sec x \right)}{\sec x+\tan x} \\
\end{align}\]
Take \[\left( \sec x \right)\] common in the numerator. Thus cancel out the like terms from the numerator and denominator.
\[\therefore \dfrac{dt}{dx}=\sec x\]
i.e. \[dt=\sec x.dx-(3)\]
Now equation (1) becomes,
\[\begin{align}
  & I=\int{\sec x.\log \left( \sec x+\tan x \right)dx} \\
 & I=\int{\left( \sec x.dx \right)\log \left( \sec x+\tan x \right)} \\
\end{align}\]
Substitute value of (2) and (3).
\[I=\int{t.dt}\]
i.e. \[I=\left( \dfrac{{{t}^{1+1}}}{1+1} \right)+C\] \[\left\{ \because {{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}+C \right\}\]
\[I=\dfrac{{{t}^{2}}}{2}+C-(4)\]
Thus let us substitute the value of t in (4).
i.e. \[t=\log \left( \sec x+\tan x \right)\]
\[\therefore I=\dfrac{{{\left[ \log \left( \sec x+\tan x \right) \right]}^{2}}}{2}+C\]
Thus we got the answer for the given expression i.e.
\[\int{\sec x.\log \left( \sec x+\tan x \right)dx}=\dfrac{{{\left[ \log \left( \sec x+\tan x \right) \right]}^{2}}}{2}+C\]

Note: We might do integration by parts, which is wrong and will yield wrong answers. Thus it is important to substitute \[\log \left( \sec x+\tan x \right)\] as t. Apply basic differentiation and integration formulas, which we should remember as well to solve these kinds of problems.