First slide

Methods of integration

Question

Let $f (x) = \frac{2 - \sin x}{2 + \cos x}$
Statement-1: The antiderivative of f is a periodic function
of period $π$ .
Statement-2: The period of the function
$g (x) = \log (2 + \cos x) + A \tan^{- 1} (B \tan x / 2) + C, A, B, C$ being contants is $2 π$ .

Difficult

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

$F (x) = 2 I_{1} + I_{2}, I_{1} = \int \frac{d x}{2 + \cos x}$ and
$\begin{array}{l} I_{2} = - \int \frac{\sin x}{2 + \cos x} d x \\ = \log (2 + \cos x) + constant. \\ I_{1} = \int \frac{2 d t}{(1 + t^{2}) (2 + \frac{1 - t^{2}}{1 + t^{2}})} = 2 \int \frac{d t}{3 + t^{2}} \\ = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan (x / 2)) + C \end{array} (t = \tan (x / 2))$
Hence $F (x) = \log (2 + \cos x) + \frac{4}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan (x / 2)) +$
$C$ which is periodic with period $2 π$ .

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