Use Integration by Parts to find $\int \cos x\cdot e^x\;dx$.
How is this done? Please, feel sorry for me and help me out. Today has been a very bad day and I can't concentrate very much on this problem because my cat died. Can you help me?
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2Is this the 7th or the 9th time that happens? – Pp.. Jan 03 '15 at 18:08
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1At least show some work you have tried before asking for help. You want to integrate a product of two functions by parts and you can't try something yourself first? – KCd Jan 03 '15 at 18:09
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2This question appears to be off-topic because it The OP appears to be guilt-tripping others into doing his or her work for them. – Ron Gordon Jan 03 '15 at 18:10
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1No I am not tricking anyone – User Jan 03 '15 at 18:13
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1@Pp.. what do you mean? – User Jan 03 '15 at 18:14
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1I know I have to use uv - Integrate u dv/dx dx – User Jan 03 '15 at 18:15
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@User It is just a pedantic joke. A reference to the saying that cats have $9$ lives (or 7 in some other cases). – Pp.. Jan 03 '15 at 18:18
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1@Pp.. Ah I see. Never heard that before – User Jan 03 '15 at 18:20
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@RonGordon With google, it takes less time to find a duplicate than to type a custom close reason. – Jan 03 '15 at 18:29
1 Answers
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In integration by parts
$$\int f(x)g'(x)\text{d}x=f(x)g(x)-\int f'(x)g(x)\text{d}x$$
there is a factor $g'(x)$ which we integrate to $g(x)$ and a factor $f(x)$ which we differentiate to $f'(x)$.
Let's take $g'(x)=e^x$ and therefore $g(x)=e^x$ and for the role of $f(x)$ we take first $\cos(x)$ and latter $\sin(x)$ for a second application of integration by parts.
$$\int\cos(x)e^x\text{d}x=\cos(x)e^x+\int\sin(x)e^x\text{d}x=\cos(x)e^x+\left(\sin(x)e^x-\int\cos(x)e^x\text{d}x\right)$$
Therefore
$$\int\cos(x)e^x\text{d}x=\frac{\cos(x)e^x+\sin(x)e^x}{2}$$
Pp..
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1Thank you for your quick answer , thank you very much for your kind support. – User Jan 03 '15 at 18:21