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I've $$I = \int{x \sin{5x}} \, dx.$$ How do I evaluate it by taking $f(x) = x$ and $g'(x) = \sin{5x}$? Thanks.

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    So where are you stuck? – 2017-01-25
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    don't know how to start and evaluate. – 2017-01-25
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    Try googling "integration by parts." You have $\int f g'=fg-\int f'g$ – 2017-01-25
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    Do you know how to apply the integration by parts formula ? – 2017-01-25
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    The OP's problem is the absence of boundaries to the integral, I guess. – 2017-01-25
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    You should now select the answer that was the most helpful, in your opinion. – 2017-01-25

4 Answers 4

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$I=\int x\sin 5x\,\mathrm dx$

$\Rightarrow f(x)=x,\,g'(x)=\sin 5x$

Use the integration by parts, i.e.,

$=\displaystyle\int u\,dv=u\,v-\int v\,du$

Here, $\displaystyle\int f(x)g'(x)dx=f(x)g(x)-\int g(x)f'(x)$

Now, $g'(x)=\sin 5x$

$\Rightarrow g(x)=\displaystyle\int\sin 5x\,\mathrm dx=\dfrac{-\cos 5x}{5}$(Choose any anti derivative of $\sin 5x$)

$\therefore\displaystyle \int\sin 5x\,\mathrm dx=\dfrac{-x\cos 5x}{5}+\int\dfrac{\cos 5x}{5}\times1 \mathrm dx$

$=\dfrac{-x\cos 5x}{5}+\dfrac{\sin 5x}{25}+C$

Source: Integration by parts

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In integrating by parts, we have

$$\int f\,dg=fg-\int g\,df$$

Now, let $f=x\implies df = dx$ and $dg=\sin(5x)\implies g=-\frac15 \cos(5x)$

Can you now proceed?

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The integral symbol without boundaries means that you looking for an antiderivative rather than an integral. The difference is the following. If $F(x)$ is an anderivative of $f$, then for all $x$ we have $F'(x)=f(x)$, the notation you are using can be written this way : $$\int f(x)\mathrm d x=F(x)+\text{Constant}.$$ The integration by parts is performed like this $$\int f(x)g'(x)\mathrm d x=f(x)g(x)-\int f'(x)g(x)\mathrm dx.$$

The general rules can be found by adding boundaries to the integral $\int\rightarrow\int_a^x$ and dropping the functions evaluated at $a$, and of course keeping the final arbitrary constant in the result. Do you understand why ?

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$$I=\int x\sin 5x\,\mathrm dx$$ $$u=x=>du=dx$$ $$dv=\sin 5x=>v=\int \sin5x\, dx$$

$$x\int\sin5x-\int(\int\sin5x \,dx)\,dx$$ by substitution : $$ \int(sin 5x)dx = \int{(sin 5x)\over5 }d(5x) = {-cos(5x)\over5} + c$$

$$x ({-\cos 5x\over5})-\int({-\cos 5x\over5})\,dx$$

by substitution : $$ \int(\cos 5x)dx = \int{(\cos 5x)\over5 }d(5x) = {\sin(5x)\over5} + c$$

$$x ({-\cos 5x\over5})+({\sin 5x\over25})+c$$