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Hello I am integrating the following function by parts:

$$\int 2x\cos(2x)\,dx$$

I let \begin{align*} u&=2x \\ \,dv &= \cos(2x) \\ v&=\frac{1}{2}\sin(2x) \end{align*}

So that brings me to $$(2x)(\frac{1}{2}\sin(2x))-2\frac{1}{2}\int \sin(2x)$$ Then$$x(\sin(2x))+\frac{1}{2}\cos(2x) + C$$

The actual question I will be doing involves definite integrals but I wanted to ensure my technique is correct.

Thanks

azimut
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user
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4 Answers4

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You forgot the $du$ at the end of your second integral. The rest is correct.

William Chang
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    I read the integration by parts formula incorrectly, I did not realize you had to multiply the integral of v by the derivative of u. Thanks – user Mar 09 '14 at 19:31
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Edit: The answer below is the answer to the original question. The question has been modified since then, and the calculation has been corrected. We keep the unmodified answer since it makes some methodological suggestions.

You can check by differentiating whether you have done an integration correctly. In this case, differentiation will show that there is an error, and will likely let you see how to fix things.

It looks as if you want to find $\int 2x\cos(2x)\,dx$. The $dx$ has been left out, not a good idea. Leaving out this sort of thing often leads to error.

Let $u=2x$ and let $dv=\cos(2x)\,dx$. Then

$du=2\,dx$ and we can take $v$ to be $-\frac{1}{2}\sin(2x)$.

So we end up with $uv-\int v\,du$.

Your $uv$ is correct. But $v\,du=-\sin(2x)$.

André Nicolas
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Using the trick here: $$\begin{align} 2x \quad &\searrow^+ \quad \cos(2x)\\ 2 \quad &\searrow^- \quad 0.5\sin(2x)\\ 0 \quad &\searrow^+ \quad -0.25\cos(2x)\\ \end{align} \\ $$ Hence $\int 2x\cos(2x) \,{\rm d}x=x\sin(2x)+0.5\cos(2x)+C$

Community
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K. Rmth
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Integration by parts formula: $$\int u \ dv = uv - \int v \ du$$ Our integral that we want to solve: $$\int 2x\cos(2x) \ dx$$ Let: $$u=2x, \ du=2 \ dx$$ $$dv=\cos(2x) \ dx$$ $$v=\int \cos(2x) \ dx$$ Use integration by substitution. Let $u=2x, \ du = 2 \ dx$ $$v=\int \dfrac{1}{2}\cos(u) \ du$$ $$v=\dfrac{1}{2}\int \cos(u) \ du$$ $$v=\dfrac{1}{2}\sin(u)$$ $$v=\dfrac{1}{2}\sin(2x)$$ Now the rest is simple. $$2x\left[\dfrac{1}{2}\sin(2x)\right]-\int \dfrac{1}{2}\sin(2x) \ dx$$ $$=x\sin(2x)-\dfrac{1}{2}\int \sin(2x) \ dx$$ $$=x\sin(2x)+\dfrac{1}{2}\cos(2x)+C$$ $$\displaystyle \therefore \int 2x\cos(2x) \ dx = x\sin(2x)+\dfrac{1}{2}\cos(2x)+C$$

Your Mistake:

You made a mistake when saying: $$\int v \ du = 2\cdot\dfrac{1}{2}\int \sin(2x) \ dx$$ The $2$ should not be there. The correct form of $\int v \ du$ is: $$\int v \ du = \dfrac{1}{2} \int \sin(2x) \ dx$$
Anonymous Computer
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