Solveeit Logo

Question

Question: Evaluate \(\int {{e^x}} (\tan x + \log (\sec x))dx\)...

Evaluate ex(tanx+log(secx))dx\int {{e^x}} (\tan x + \log (\sec x))dx

Explanation

Solution

Here in this question we will first solve the bracket and multiply the given terms then we will apply integration by parts. To solve this question we required some integration and differentiation formulas which are mentioned below:-
exdx=ex\int {{e^x}} dx = {e^x}
ddx(secx)=secxtanx\dfrac{d}{{dx}}(\sec x) = \sec x\tan x
ddxlogx=1x\dfrac{d}{{dx}}\log x = \dfrac{1}{x}
Integration by parts is done when integration of two products have to be find out here is the formula where U is taken as first function and V is taken as second function U.Vdx=UVdx[ddxUVdx]\int {U.V} dx = U\int {Vdx - [\int {\dfrac{d}{{dx}}U} } \int {Vdx} ]

Complete step-by-step answer:
We have to evaluate this integral given ex(tanx+log(secx))dx\int {{e^x}} (\tan x + \log (\sec x))dx so first of all we will multiply exponential function inside with other functions.
extanxdx+exlog(secx)dx\Rightarrow \int {{e^x}} \tan xdx + {e^x}\log (\sec x)dx
Now we will apply integration by parts U.Vdx=UVdx[ddxUVdx]\int {U.V} dx = U\int {Vdx - [\int {\dfrac{d}{{dx}}U} } \int {Vdx} ] in exlog(secx){e^x}\log (\sec x) where U will be log(secx)\log (\sec x)and V will be ex{e^x}
extanxdx+log(secx)exdx[ddxlog(secx)exdx]\Rightarrow \int {{e^x}} \tan xdx + \log (\sec x)\int {{e^x}dx - [\int {\dfrac{d}{{dx}}\log (\sec x)} } \int {{e^x}dx} ]
Now we will do integration of exponential function and differentiation of log (sec x) function.
extanxdx+log(secx)(ex)1secx×secxtanx(ex)dx\Rightarrow \int {{e^x}} \tan xdx + \log (\sec x)({e^x}) - \int {\dfrac{1}{{\sec x}} \times \sec x\tan x({e^x})dx}
Here we have applied chain rule where we first did differentiation of log x which is 1/x then we have done differentiation of sec x which is sec x tan x. Now we will just cancel sec x from numerator and denominator.
extanxdx+log(secx)(ex)tanx(ex)dx\Rightarrow \int {{e^x}} \tan xdx + \log (\sec x)({e^x}) - \int {\tan x({e^x})} dx
Now we will simply cancel equal and opposite integral terms form the equation.
log(secx)(ex)+C\Rightarrow \log (\sec x)({e^x}) + C (Where C is the constant)

Therefore integral value of ex(tanx+log(secx))dx\int {{e^x}} (\tan x + \log (\sec x))dx is log(secx)(ex)+C\log (\sec x)({e^x}) + C

Note: Students may get confused while applying integration by parts in which we have to choose which function we should take first and which to take second for this there is a short trick of ILATE which is briefly explained below.
ILATE is a short trick which helps to find out the order for applying integration by parts. Priority is given from top to bottom.
I = Integer
L = logarithmic function
A= algebra
T= trigonometric function
E= exponential function