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Question Number 53472 by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

$s o u r c e . . . y o u t u b e . . .$
Commented by malwaan last updated on 23/Jan/19

$can you solve it sir ?$ $please$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

	$I = \int_{- a}^{a} \frac{{c o s}^{b} x + 1}{c^{x} + 1} d x$ $a = 12345678910 π m u l t i p l e o f 2 π$ $b = 123456789 o d d n u m b e r$ $c = 201620172018 e v e n n u m b e r$ $n o w l e t t = - x d x = - d t$ $\int_{a}^{- a} \frac{{c o s}^{b} (- t) + 1}{c^{- t} + 1} (- d t)$ $\int_{a}^{- a} \frac{{c o s}^{b} t + 1}{\frac{1}{c^{t}} + 1} \times - d t$ $\int_{- a}^{a} \frac{c^{t} ({c o s}^{b} t + 1)}{1 + c^{t}} d t$ $n o w \int_{- a}^{a} f (x) d x = \int_{- a}^{a} f (t) d t$ $2 I = \int_{- a}^{a} \frac{{c o s}^{b} x + 1}{c^{x} + 1} d x + \int_{- a}^{a} \frac{c^{t} ({c o s}^{b} t + 1)}{1 + c^{t}} d t$ $2 I = \int_{- a}^{a} (1 + {c o s}^{b} t) d t$ $I = \frac{1}{2} \int_{- a}^{a} d t + \frac{1}{2} \int_{- a}^{a} {c o s}^{b} t d t$ $I = \frac{1}{2} \times 2 a + I_{1}$ $I = a + I_{1}$ $[\frac{1}{2} \int_{- a}^{a} {c o s}^{b} t d t = 0]$ $s o a n s w d r i s I = a = 12345678910 π$
	Commented by malwaan last updated on 24/Jan/19

	$thank you for your time$
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