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 $\tan x={{\sec }^{2}}x$.

& \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{aligned}$$ Take $$\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{aligned} & 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{aligned}$$ Substitute value of (2) and (3). $$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.