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Q20:
Integrate the functions in Exercises 1 to 23. 20. $\frac{2+\sin 2x}{1+\cos 2x} e^x$

Solution :

Step 1: Trigonometric Simplification

The integrand is $I = \int \frac{2+\sin 2x}{1+\cos 2x} e^x \, dx$. We simplify the rational expression using double-angle identities [Per Trigonometric Identities]:

  • $\sin 2x = 2 \sin x \cos x$
  • $1 + \cos 2x = 2 \cos^2 x$

Substituting these into the expression:

$f(x) = \frac{2 + 2 \sin x \cos x}{2 \cos^2 x}$

Divide the numerator by the denominator:

$f(x) = \frac{2}{2 \cos^2 x} + \frac{2 \sin x \cos x}{2 \cos^2 x} = \sec^2 x + \tan x$

Step 2: Structuring the Integral

The integral now becomes:

$I = \int e^x (\tan x + \sec^2 x) \, dx$

We observe that this fits the standard integration form [Per the Integration by Parts theorem for the form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$]:

  • Let $f(x) = \tan x$
  • Then $f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$

Step 3: Execution of the Standard Result

Applying the formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$:

$I = e^x \tan x + C$ x y f(x) = e^x tan(x)

Final Solution: $e^x \tan x + C$


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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