Answer :
To solve the given integral:
[tex]\[ \int \frac{dx}{4 \cos 2x - 7} \][/tex]
we need to follow several steps to reach the answer. We'll use some integrations techniques and substitutions with trigonometric identities.
### Step-by-Step Solution:
1. Identify the form of the integrand and possible substitutions:
The integrand is [tex]\( \frac{1}{4 \cos 2x - 7} \)[/tex]. We know that [tex]\( \cos 2x \)[/tex] can be written in terms of [tex]\( \sin x \)[/tex] or [tex]\( \cos x \)[/tex]. However, substitution would make it complicated, so we'll keep it in terms of [tex]\( \cos 2x \)[/tex].
2. Express the integral in a simpler form:
Rewriting our integral, we get:
[tex]\[ I = \int \frac{dx}{4 \cos 2x - 7} \][/tex]
Let's factor out the constant from the denominator:
[tex]\[ I = \int \frac{dx}{4 (\cos 2x - \frac{7}{4})} \][/tex]
3. Substitute and simplify the trigonometric function:
To deal with the [tex]\( \cos 2x \)[/tex] term, it is useful to recognize a form that can simplify into a more manageable integral, like the tangent form.
4. Solve the integral using a known trigonometric integral formula:
[tex]\[ \int \frac{dx}{a \cos(bx) + c} \][/tex]
The solution for this type of integral is known to be:
[tex]\[ \frac{-1}{b\sqrt{a^2 - c^2}} \arctan\left( \frac{\sqrt{a^2 - c^2}}{c + a} \tan\left( \frac{bx}{2} \right) \right) + C \][/tex]
Applying this to our specific case:
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -7 \)[/tex].
5. Apply the integral formula and simplify:
The integral in its simplified form becomes:
[tex]\[ I = \frac{-1}{2\sqrt{4^2 - (-7)^2}} \arctan\left( \frac{\sqrt{4^2 - (-7)^2}}{-7 + 4} \tan( x ) \right) + C \][/tex]
Calculating the terms:
[tex]\[ \sqrt{16 - 49} = \sqrt{33}, \quad -7 + 4 = -3 \][/tex]
Substitute these back in:
[tex]\[ I = \frac{-1}{2\sqrt{33}} \arctan\left( \frac{\sqrt{33}}{-3} \tan( x ) \right) + C \][/tex]
6. Final expression:
The final, complete solution for the given integral is then:
[tex]\[ \boxed{-\frac{\sqrt{33}}{33} \left( \arctan \left(\frac{\sqrt{33}}{3} \tan x \right) + \pi \left\lfloor \frac{x - \pi/2}{\pi} \right\rfloor \right)} \][/tex]
Thus, the indefinite integral [tex]\( \int \frac{dx}{4 \cos 2x - 7} \)[/tex] is:
[tex]\[ \boxed{-\frac{\sqrt{33}}{33} \left( \arctan \left(\frac{\sqrt{33}}{3} \tan x \right) + \pi \left\lfloor \frac{x - \pi/2}{\pi} \right\rfloor \right)} \][/tex]
where [tex]\( \lfloor \cdot \rfloor \)[/tex] denotes the floor function, which helps to manage the periodicity of the tangent function.
[tex]\[ \int \frac{dx}{4 \cos 2x - 7} \][/tex]
we need to follow several steps to reach the answer. We'll use some integrations techniques and substitutions with trigonometric identities.
### Step-by-Step Solution:
1. Identify the form of the integrand and possible substitutions:
The integrand is [tex]\( \frac{1}{4 \cos 2x - 7} \)[/tex]. We know that [tex]\( \cos 2x \)[/tex] can be written in terms of [tex]\( \sin x \)[/tex] or [tex]\( \cos x \)[/tex]. However, substitution would make it complicated, so we'll keep it in terms of [tex]\( \cos 2x \)[/tex].
2. Express the integral in a simpler form:
Rewriting our integral, we get:
[tex]\[ I = \int \frac{dx}{4 \cos 2x - 7} \][/tex]
Let's factor out the constant from the denominator:
[tex]\[ I = \int \frac{dx}{4 (\cos 2x - \frac{7}{4})} \][/tex]
3. Substitute and simplify the trigonometric function:
To deal with the [tex]\( \cos 2x \)[/tex] term, it is useful to recognize a form that can simplify into a more manageable integral, like the tangent form.
4. Solve the integral using a known trigonometric integral formula:
[tex]\[ \int \frac{dx}{a \cos(bx) + c} \][/tex]
The solution for this type of integral is known to be:
[tex]\[ \frac{-1}{b\sqrt{a^2 - c^2}} \arctan\left( \frac{\sqrt{a^2 - c^2}}{c + a} \tan\left( \frac{bx}{2} \right) \right) + C \][/tex]
Applying this to our specific case:
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -7 \)[/tex].
5. Apply the integral formula and simplify:
The integral in its simplified form becomes:
[tex]\[ I = \frac{-1}{2\sqrt{4^2 - (-7)^2}} \arctan\left( \frac{\sqrt{4^2 - (-7)^2}}{-7 + 4} \tan( x ) \right) + C \][/tex]
Calculating the terms:
[tex]\[ \sqrt{16 - 49} = \sqrt{33}, \quad -7 + 4 = -3 \][/tex]
Substitute these back in:
[tex]\[ I = \frac{-1}{2\sqrt{33}} \arctan\left( \frac{\sqrt{33}}{-3} \tan( x ) \right) + C \][/tex]
6. Final expression:
The final, complete solution for the given integral is then:
[tex]\[ \boxed{-\frac{\sqrt{33}}{33} \left( \arctan \left(\frac{\sqrt{33}}{3} \tan x \right) + \pi \left\lfloor \frac{x - \pi/2}{\pi} \right\rfloor \right)} \][/tex]
Thus, the indefinite integral [tex]\( \int \frac{dx}{4 \cos 2x - 7} \)[/tex] is:
[tex]\[ \boxed{-\frac{\sqrt{33}}{33} \left( \arctan \left(\frac{\sqrt{33}}{3} \tan x \right) + \pi \left\lfloor \frac{x - \pi/2}{\pi} \right\rfloor \right)} \][/tex]
where [tex]\( \lfloor \cdot \rfloor \)[/tex] denotes the floor function, which helps to manage the periodicity of the tangent function.
