∫e^x sin x dx = −e^x cos x + e^x sin x − ∫ e^x sin x dx


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Question Number 101212 by rb222 last updated on 01/Jul/20

$\int e^{x} \sin x d x = - e^{x} c o s x + e^{x} s i n x - \int e^{x} \sin x d x$
Commented by Dwaipayan Shikari last updated on 01/Jul/20

$\int e^{x} s i n x d x = e^{x} \int s i n x d x + \int e^{x} c o s x d x$ $I = - e^{x} c o s x + e^{x} s i n x - \int e^{x} s i n x d x$ $I = e^{x} (s i n x - c o s x) - I$ $2 I = e^{x} (s i n x - c o s x) + C$ $I = \frac{e^{x} (s i n x - c o s x)}{2} + C o n s t a n t (o r \frac{C}{2})$
	Answered by smridha last updated on 01/Jul/20
	$Im∫e^((1+i)x) dx=Im[(e^((1+i)x) /((1+i)))]+c =(1/2)Im[(1−i)e^x (cosx+isinx)]+c =(1/2)Im[e^x {(cosx+sinx)+i(sinx−cosx)}]+c =(e^x /2)(sinx−cosx)+c ans or =(e^x /(√2))(sinxcos(𝛑/4)−cosx.sin(𝛑/4))+c =(e^x /(√2))sin(x−(𝛑/4))+c ans or =(e^x /(√2))cos[(𝛑/2)+(x−(𝛑/4))]+c =(e^x /(√2))cos[x+(𝛑/4)]+c ans let I=∫e^x sinx=e^x sinx−∫e^x cosx =e^x sinx−e^x cosx−∫e^x sinx =e^x (sinx−cosx)−I+k 2I=e^x (sinx−cosx)+k so I=(e^x /2)(sinx−cosx)+c [c=(k/2)=constant]$
	$I m \int e^{(1 + i) x} d x = I m [\frac{e^{(1 + i) x}}{(1 + i)}] + c$ $= \frac{1}{2} I m [(1 - i) e^{x} (c o s x + i s i n x)] + c$ $= \frac{1}{2} I m [e^{x} {(c o s x + s i n x) + i (s i n x - c o s x)}] + c$ $= \frac{e^{x}}{2} (s i n x - c o s x) + c a n s$ $o r = \frac{e^{x}}{\sqrt{2}} (s i n x c o s \frac{π}{4} - c o s x . s i n \frac{π}{4}) + c$ $= \frac{e^{x}}{\sqrt{2}} s i n (x - \frac{π}{4}) + c a n s$ $o r = \frac{e^{x}}{\sqrt{2}} c o s [\frac{π}{2} + (x - \frac{π}{4})] + c$ $= \frac{e^{x}}{\sqrt{2}} c o s [x + \frac{π}{4}] + c a n s$ $l e t$ $I = \int e^{x} s i n x = e^{x} s i n x - \int e^{x} c o s x$ $= e^{x} s i n x - e^{x} c o s x - \int e^{x} s i n x$ $= e^{x} (s i n x - c o s x) - I + k$ $2 I = e^{x} (s i n x - c o s x) + k$ $s o I = \frac{e^{x}}{2} (s i n x - c o s x) + c [c = \frac{k}{2} = c o n s t a n t]$
	Commented by smridha last updated on 01/Jul/20

	$w h o a r e y o u m a n ? ? w e l l c o p i e d!!$
	Answered by MJS last updated on 01/Jul/20

	$why so complicated ?$ ${it}^{'} s simply 2 times by parts$ $I = \int e^{x} \sin x d x =$ $u^{'} = u = e^{x}$ $v = \sin x \to v^{'} = \cos x$ $= e^{x} \sin x - \int e^{x} \cos x d x =$ $u^{'} = u = e^{x}$ $v = \cos x \to v^{'} = - \sin x$ $= e^{x} \sin x - (e^{x} \cos x - \int - e^{x} \sin x d x) =$ $= e^{x} \sin x - e^{x} \cos x - \int e^{x} \sin x d x$ $now we have$ $I = e^{x} \sin x - e^{x} \cos x - I$ $2 I = e^{x} \sin x - e^{x} \cos x$ $I = \frac{e^{x} \sin x - e^{x} \cos x}{2} + C$
	Commented by smridha last updated on 02/Jul/20
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