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Q11:
Integrate the functions in Exercises 1 to 23. 11. $\frac{1}{\cos(x+a)\cos(x+b)}$

Solution :

Step 1: Algebraic Manipulation of the Integrand

To integrate $I = \int \frac{dx}{\cos(x+a)\cos(x+b)}$, we must express the numerator in a form that allows for the simplification of the trigonometric products in the denominator. We utilize the constant $\sin(b-a)$ as a scaling factor [By the property of linear independence of trigonometric functions].

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\cos(x+a)\cos(x+b)} dx$

Step 2: Trigonometric Substitution

We rewrite the numerator $(b-a)$ as $(x+b) - (x+a)$ to match the arguments present in the denominator:

$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x+b) - (x+a))}{\cos(x+a)\cos(x+b)} dx$

Step 3: Expansion via Sine Addition Formula

Applying the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ [Per the Compound Angle Formula], where $A = (x+b)$ and $B = (x+a)$:

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(x+b)\cos(x+a) - \cos(x+b)\sin(x+a)}{\cos(x+a)\cos(x+b)} dx$

Step 4: Separating the Integral

Distribute the denominator across the terms in the numerator:

$I = \frac{1}{\sin(b-a)} \left[ \int \frac{\sin(x+b)\cos(x+a)}{\cos(x+a)\cos(x+b)} dx - \int \frac{\cos(x+b)\sin(x+a)}{\cos(x+a)\cos(x+b)} dx \right]$

$I = \frac{1}{\sin(b-a)} \left[ \int \frac{\sin(x+b)}{\cos(x+b)} dx - \int \frac{\sin(x+a)}{\cos(x+a)} dx \right]$

$I = \frac{1}{\sin(b-a)} \left[ \int \tan(x+b) dx - \int \tan(x+a) dx \right]$

Step 5: Final Integration

Applying the standard integral result $\int \tan(u) du = \ln|\sec(u)| + C$ or $-\ln|\cos(u)| + C$ [Per the Fundamental Theorem of Calculus]:

$I = \frac{1}{\sin(b-a)} \left[ \ln|\sec(x+b)| - \ln|\sec(x+a)| \right] + C$

Using the logarithmic property $\ln(m) - \ln(n) = \ln(\frac{m}{n})$:

$I = \frac{1}{\sin(b-a)} \ln \left| \frac{\sec(x+b)}{\sec(x+a)} \right| + C$

Substituting $\sec(\theta) = \frac{1}{\cos(\theta)}$:

$I = \frac{1}{\sin(b-a)} \ln \left| \frac{\cos(x+a)}{\cos(x+b)} \right| + C$

Integral of: 1 / [cos(x+a)cos(x+b)] Method: Trigonometric Identity Substitution Result: (1 / sin(b-a)) * ln |cos(x+a)/cos(x+b)| + C

Final Solution: $\frac{1}{\sin(b-a)} \ln \left| \frac{\cos(x+a)}{\cos(x+b)} \right| + C$


More Questions from Class 12 Mathematics Integrals Miscellaneous Exercise on Chapter 7


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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