Question Statement Evaluate the integral:
β« 3 sin β‘ x + cos β‘ x , d x \int \frac{\sqrt{3}}{\sin x + \cos x} , dx β« sin x + cos x 3 β β , d x Background and Explanation To solve this integral, we need to simplify the expression in the denominator. We can use trigonometric identities to rewrite the terms and make the integral easier to solve. Specifically, the sum of sine and cosine functions can be rewritten using a standard identity. This will help us transform the integral into a more familiar form, which can then be solved using basic integration techniques.
Solution Hereβs the step-by-step solution:
Factor Out Constants :
First, we factor out the constant 1 2 \frac{1}{\sqrt{2}} 2 β 1 β
β« 3 sin β‘ x + cos β‘ x , d x = 1 2 β« 2 sin β‘ x + cos β‘ x 2 , d x \int \frac{\sqrt{3}}{\sin x + \cos x} , dx = \frac{1}{\sqrt{2}} \int \frac{\sqrt{2}}{\frac{\sin x + \cos x}{\sqrt{2}}} , dx β« sin x + cos x 3 β β , d x = 2 β 1 β β« 2 β s i n x + c o s x β 2 β β , d x
Use Trigonometric Identity :
Next, we apply the well-known trigonometric identity for sin β‘ ( x + Ο 4 ) \sin \left(x + \frac{\pi}{4}\right) sin ( x + 4 Ο β )
sin β‘ x cos β‘ Ο 4 + cos β‘ x sin β‘ Ο 4 = sin β‘ ( x + Ο 4 ) \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \sin \left(x + \frac{\pi}{4}\right) sin x cos 4 Ο β + cos x sin 4 Ο β = sin ( x + 4 Ο β ) Using this identity, we can rewrite the integral as:
β« 1 sin β‘ x sin β‘ Ο 4 + cos β‘ x cos β‘ Ο 4 , d x = β« 1 cos β‘ ( x β Ο 4 ) , d x \int \frac{1}{\sin x \sin \frac{\pi}{4} + \cos x \cos \frac{\pi}{4}} , dx = \int \frac{1}{\cos \left(x - \frac{\pi}{4}\right)} , dx β« sin x sin 4 Ο β + cos x cos 4 Ο β 1 β , d x = β« cos ( x β 4 Ο β ) 1 β , d x
Rewrite Using the Secant Function :
The expression 1 cos β‘ ( x β Ο 4 ) \frac{1}{\cos \left(x - \frac{\pi}{4}\right)} c o s ( x β 4 Ο β ) 1 β sec β‘ ( x β Ο 4 ) \sec \left(x - \frac{\pi}{4}\right) sec ( x β 4 Ο β )
β« sec β‘ ( x β Ο 4 ) , d x \int \sec \left(x - \frac{\pi}{4}\right) , dx β« sec ( x β 4 Ο β ) , d x
Integrate Using the Standard Formula :
We now use the standard integral formula for sec β‘ x \sec x sec x
β« sec β‘ x , d x = ln β‘ β£ sec β‘ x + tan β‘ x β£ + C \int \sec x , dx = \ln \left| \sec x + \tan x \right| + C β« sec x , d x = ln β£ sec x + tan x β£ + C Applying this formula to our integral:
ln β‘ sec β‘ ( x β Ο 4 ) + tan β‘ ( x β Ο 4 ) + C \ln \sec \left(x - \frac{\pi}{4}\right) + \tan \left(x - \frac{\pi}{4}\right) + C ln sec ( x β 4 Ο β ) + tan ( x β 4 Ο β ) + C
Trigonometric Identity :
sin β‘ ( x + Ο 4 ) = sin β‘ x cos β‘ Ο 4 + cos β‘ x sin β‘ Ο 4 \sin \left(x + \frac{\pi}{4}\right) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} sin ( x + 4 Ο β ) = sin x cos 4 Ο β + cos x sin 4 Ο β
Secant Function :
sec β‘ ( x ) = 1 cos β‘ ( x ) \sec(x) = \frac{1}{\cos(x)} sec ( x ) = c o s ( x ) 1 β
Standard Integral :
β« sec β‘ x , d x = ln β‘ β£ sec β‘ x + tan β‘ x β£ + C \int \sec x , dx = \ln \left| \sec x + \tan x \right| + C β« sec x , d x = ln β£ sec x + tan x β£ + C
Summary of Steps
Factor out the constant 1 2 \frac{1}{\sqrt{2}} 2 β 1 β
Apply the trigonometric identity to simplify the denominator.
Rewrite the integral as β« sec β‘ ( x β Ο 4 ) , d x \int \sec(x - \frac{\pi}{4}) , dx β« sec ( x β 4 Ο β ) , d x
Use the standard integral formula for sec β‘ x \sec x sec x
The final result is:
ln β‘ sec β‘ ( x β Ο 4 ) + tan β‘ ( x β Ο 4 ) + C \ln \sec \left(x - \frac{\pi}{4}\right) + \tan \left(x - \frac{\pi}{4}\right) + C ln sec ( x β 4 Ο β ) + tan ( x β 4 Ο β ) + C 
