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Question Number 142120 by bekzodjumayev last updated on 26/May/21

Commented by bekzodjumayev last updated on 26/May/21

$H e l p$
Commented by Avijit007 last updated on 26/May/21

$g o t i t$
	Answered by som(math1967) last updated on 26/May/21
	$∫((e^x dx)/(2cos^2 (x/2)))+∫((2sin(x/2)cos(x/2)e^x dx)/(2cos^2 (x/2))) (1/2)∫sec^2 (x/2)e^x dx +∫tan(x/2)e^x dx (1/2)∫sec^2 (x/2)e^x dx +tan(x/2)∫e^x dx −∫{(d/dx)tan(x/2)∫e^x dx}dx (1/2)∫sec^2 (x/2)e^x dx +e^x tan(x/2)−(1/2)∫sec^2 (x/2)e^x dx e^x tan(x/2) +C$
	$\int \frac{e^{x} d x}{2 {c o s}^{2} \frac{x}{2}} + \int \frac{2 s i n \frac{x}{2} c o s \frac{x}{2} e^{x} d x}{2 {c o s}^{2} \frac{x}{2}}$ $\frac{1}{2} \int {s e c}^{2} \frac{x}{2} e^{x} d x + \int t a n \frac{x}{2} e^{x} d x$ $\frac{1}{2} \int {s e c}^{2} \frac{x}{2} e^{x} d x + t a n \frac{x}{2} \int e^{x} d x$ $- \int {\frac{d}{d x} t a n \frac{x}{2} \int e^{x} d x} d x$ $\frac{1}{2} \int {s e c}^{2} \frac{x}{2} e^{x} d x + e^{x} t a n \frac{x}{2} - \frac{1}{2} \int {s e c}^{2} \frac{x}{2} e^{x} d x$ $e^{x} t a n \frac{x}{2} + C$
	Commented by bekzodjumayev last updated on 26/May/21

	$T h a n k y o u$
	Answered by Avijit007 last updated on 26/May/21
	$The given integral is, ∫((1+sin x)/(1+cos x))e^x .dx =∫[e^x {(1/(1+cosx))+((sinx)/(1+sinx))}].dx =∫[e^x {(1/(2cos^2 ((x/2))))+((2sin((x/2))cos((x/2)))/(2cos^2 ((x/2))))].dx =∫[e^x {(1/2)sec^2 ((x/2))+tan((x/2))].dx ⇓ ⇓ f ′(x) f(x) ∵ we know that, ∫e^x {f ′(x)+ f(x)} ⇒e^x f(x)+c = e^x .tan((x/2))+c {ANS} by Avijit$
	$T h e g i v e n i n t e g r a l i s,$ $\int \frac{1 + s i n x}{1 + c o s x} e^{x} . d x$ $= \int [e^{x} {\frac{1}{1 + c o s x} + \frac{s i n x}{1 + s i n x}}] . d x$ $= \int [e^{x} {\frac{1}{2 {c o s}^{2} (\frac{x}{2})} + \frac{2 s i n (\frac{x}{2}) c o s (\frac{x}{2})}{2 {c o s}^{2} (\frac{x}{2})}] . d x$ $= \int [e^{x} {\frac{1}{2} {s e c}^{2} (\frac{x}{2}) + t a n (\frac{x}{2})] . d x$ $⇓ ⇓$ $f^{'} (x) f (x)$ $∵ w e k n o w t h a t, \int e^{x} {f^{'} (x) + f (x)}$ $\Rightarrow e^{x} f (x) + c$ $= e^{x} . t a n (\frac{x}{2}) + c {ANS}$ $b y A v i j i t$
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