Can't understand how to solve this math: use integration by parts to evaluate this integrals:
$$\int x\sin(2x + 1) \,dx$$
can any one solve this so i can understand how to do this! Thanks :)
$\endgroup$ 24 Answers
$\begingroup$$$\int x \sin (2x+1)\;dx$$ parts by integration:$\int u(x)\cdot v(x) \;dx=u\int v(x)\;dx-\int \dfrac {d}{dx}\,u(x)[\int v(x)\,dx]\;dx$ Here one question is that which function should be taken as $u$ and $v$.so I learnt that use "ILATE". This is the sequnce to choose $u$ and $v$.
I=inverse function,L=log,A=airthematic,T=trigonometry,E=exponential. This will help most of the time in integration by parts. $$x\int \sin (2x+1)-\int \dfrac {d}{dx}x \int \sin (2x+1)\; dx\;dx$$ $$x(-\dfrac {\cos (2x+1)}{2})-\int 1\cdot (-\dfrac {\cos (2x+1)}{2})\; dx$$ $$\dfrac {-x\cos (2x+1)}{2}+\dfrac {\sin (2x+1)}{4}+C$$
$\endgroup$ 2 $\begingroup$$\int uv'=uv-\int u'v$. Choose $u(x):=x$, $v'(x):=\sin(2x+1)$. Then $u'(x)=1$ and $v(x)=-\frac{\cos(2x+1)}{2}$. So $$ \int x\sin(2x+1)\,dx=-x\frac{\cos(2x+1)}{2}+\int\frac{\cos(2x+1)}{2}\,dx=-x\frac{\cos(2x+1)}{2}+\frac{\sin(2x+1)}{4}+C. $$
$\endgroup$ $\begingroup$I'm accustomed to seeing integration by parts expressed like this: $$ \int u\,dv = uv - \int v\,du. $$ The idea is that $u$ is something you can easily differentiate and $dv$ is something you can easily antidifferentiate and $\int v\,du$ is somehow more tractable than $\int u\,dv$.
$$ \int x\Big(\sin(2x+1)\,dx\Big) = \int x\,dv = xv - \int v\,dx. $$ We have $dv = \sin(2x+1)$, so $v=\frac{-1}{2}\cos(2x+1)$. So $$ xv - \int v\,dx = \frac{-1}{2}x\cos(2x+1) - \int \frac{-1}{2} \cos(2x+1)\,dx. $$ $$ =\frac{-1}{2}x\cos(2x+1) +\frac12 \int \cos(2x+1)\,dx $$ $$ =\frac{-1}{2}x\cos(2x+1) +\frac14 \sin(2x+1) + C. $$
$\endgroup$ $\begingroup$The idea behind integrating by parts is the following. If $u,v$ are functions of $x$ then using the chain rule we know that:
$$\frac{d(uv)}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}$$
Thus, integrating throughout with respect to $x$, we get:
$$uv = \int{\frac{d(u)}{dx}v dx} + \int{u\frac{dv}{dx}}$$
Or in other words:
$$\int{\frac{du}{dx}v dx} = uv - \int{u\frac{dv}{dx}}$$
So, on the left is an integral and on the right is a product of two functions and another integral. So, when we have to do an integral using integration by parts the strategy is to choose $u$ and $v$ carefully so that:
Step 1: We choose what part of the integrand represents $\frac{du}{dx}$ such that we can identify the function $u$ which when differentiated with respect to $x$ gets us that part of the integrand
and
Step 2: Choose which part of the integrand represents $v$.
The ultimate goal is to reduce the integral (the one on the left) we have to solve by another integral (the one on the right) which hopefully is easier to solve given a judicious choice of $u$ and $v$.
Let us examine the above strategy in your example. We have:
$$\int{x\sin(2x+1)dx}$$
Step 1: Suppose, we try $u=-0.5\cos(2x+1)$
$$\frac{d[-0.5 \cos(2x+1)]}{dx}=\sin(2x+1)$$
and
Step 2: And set $v=x$
$$\frac{d(x)}{dx}=1$$
Thus, setting $u=-0.5\cos(2x+1)$ and $v=x$ may get us a simpler integral to work with. So, we get:
$$\int{x\sin(2x+1)dx} = x[-0.5\cos(2x+1)]-\int{-0.5\cos(2x+1)dx}$$
The integral on the right $\int{-0.5\cos(2x+1)dx}$ is indeed simpler and easy to solve.
Hopefully, the above will help with not just this particluar integral but any future integrals you need to do using integration by parts.
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