একাধিকবার খণ্ডায়ন সূত্রের মাধ্যমে যোগজীকরণ

In the previous section, we learned about integration by parts, which states that for continuously differentiable functions $u(x)$ and $v(x)$ we have $$\int u\, dv=uv-\int v\, du.$$ This time, let's continue with a more elaborate example:

Let's try integrating by parts: we let $u_1(x)=x^2$ and $dv_1=\cos{x}\, dx$. This gets us $$\frac{du_1}{dx}=2x\Longrightarrow du_1=2x\, dx$$ and $$v_1(x)=\int\, dv_1=\int \cos{x}\, dx=\sin{x}.$$ So, we have $$\int x^2\cos{x}=\int u_1\, dv_1=u_1v_1-\int v_1\, du_1=x^2\sin{x}-\int \sin{x}\cdot (2x\, dx)$$ $$=x^2\sin{x}-2\int x\sin{x}\, dx.$$ Okay, but can we simplify that second integral? In fact, we can by using the same process! This time, we'll let $u_2(x)=x$ and $dv_2=\sin{x}\, dx$. Then $$\frac{du_2}{dx} =1\Longrightarrow du_2=dx$$ and $$v_2(x)=\int\, dv_2=\int \sin{x}\, dx=-\cos{x}.$$ So, this gets us $$\int x\sin{x}=\int u_2\, dv_2=u_2v_2-\int v_2\, du_2=x(-\cos{x})-\int (-\cos{x})\, dx$$ $$=-x\cos{x}+\int \cos{x}\, dx=-x\cos{x}+\sin{x}+c.$$ But remember, this isn't our final answer! We need to plug this integral back into our original expression to get $$\int x^2\cos{x}=x^2\sin{x}-2\int x\sin{x}\, dx=x^2\sin{x}-2\left(-x\cos{x}+\sin{x}+c\right)$$ $$=x^2\sin{x}+2x\cos{x}-2\sin{x}+C,$$ where $C=-2c$ is just a new constant. This is our final answer.

As in this problem, you can see that a problem might require us to use this method more than once! With this in mind, try this next exercise:

Let's start with the same approach as usual: we set $u_1(x)=\sin{x}$ and $dv_1=e^x\, dx$. So, $$\frac{du_1}{dx}=\cos{x}\Longrightarrow du_1=\cos{x}\, dx$$ and $$v_1(x)=\int v_1\, dx=\int e^x\, dx=e^x.$$ So, it follows that $$\int e^x\sin{x}=\int u_1\, dv_1=u_1v_1-\int v_1\, du_1=e^x\sin{x}-\int e^x\cos{x}\, dx.$$ Just like before, we have another integral that we need to solve by using integration by parts! This time, we let $u_2(x)=\cos{x}$ and $dv_2=e^x\, dx$. This gets us $$\frac{du_2}{dx}=-\sin{x}\Longrightarrow du_2=-\sin{x}\, dx$$ and $$v_2(x)=\int \, dv_2=\int e^x\, dx=e^x.$$ So, by integration by parts, $$\int e^x\cos{x}\, dx=\int u_2\, dv_2=u_2v_2-\int v_2\, du_2=\cos{x}e^x-\int e^x(-\sin{x}\, dx)$$ $$=e^x\cos{x}+\int e^x\sin{x}\, dx.$$ Darn, another complicated integral to compute! Well, not quite. Here's when we take a ``shortcut''. Let's substitute this integral into our original expression: $$\int e^x\sin{x}=e^x\sin{x}-\int e^x\cos{x}\, dx\\=e^x\sin{x}-\left( e^x\cos{x}+\int e^x\sin{x}\, dx\right)$$ $$=e^x\sin{x}-e^x\cos{x}-\int e^x\sin{x}\, dx.$$ Now, do you notice something similar in both sides the equation?

Aha! The left and right hand sides both have the integral $\int e^x\sin{x}\, dx$. So, let's ``combine like terms'' and add the integral to both sides to get $$2\int e^x\sin{x}\, dx=e^x\sin{x}-e^x\cos{x}.$$ Therefore, we can solve for the integral to get $$\int e^x\sin{x}\, dx=\frac{e^x\sin{x}-e^x\cos{x}}{2}.$$ This brings us to the end of the problem!

So, these past examples show that sometimes, we need to apply integration by parts more than once, and in unusual ways! But these methods also demonstrate just how useful this technique can be.

In some cases, integration by parts must be used on an integral multiple times in succession. One must be able to recognize when this is necessary, and how to clean up the resulting mess. Here we will walk through a common example.

Integrate $\int{e^x\cos(x)dx}$.

We can integrate by parts once, initially integrating the $e^x$ "part", giving us

$\int{e^x\cos(x)dx} = e^x\cos(x) - \int{e^x\frac{d}{dx}(\cos(x))dx} \Rightarrow$

$\int{e^x\cos(x)dx} = e^x\cos(x) + \int{e^x\sin(x)dx}$

We now have a new integral! This one also needs to be integrated by parts, in the same fashion as the first:

$\int{e^x\cos(x)dx} = e^x\cos(x) + e^x\sin(x) - \int{e^x\cos(x)dx}$

Now solve for the original integral (pay close attention to this step):

$2\int{e^x\cos(x)dx} = e^x\cos(x) + e^x\sin(x) \Rightarrow$

$\int{e^x\cos(x)dx} = \frac{e^x\cos(x) + e^x\sin(x)}{2} + C$

(Video) Integration by Parts: Looping

by PatrickJMT

Introduction

Sometimes, while doing integration by parts, you loop back around and your answer contains the very integral you're trying to solve. These types of problems are usually identified with a combination of trig functions (because they loop) and exponential functions (because they also loop).

In this video, Patrick shows you the trick on how to solve these.

Tip: a lot of professors love to throw one of these in on the test, so be careful!

Example

Let's look at an example in order to demonstrate everything above. The example he looks at is: $$\int e^{2x}\sin x\,dx$$

Start by applying by parts. LIATE tells us that $u$ should be the trig portion. $$\begin{array}{cc} \color{red}{u=\sin x}&v=\frac12e^{2x}\\ \downarrow&\uparrow\\ du=\cos x\,dx&\color{red}{dv=e^{2x}\,dx} \end{array}$$

$$\begin{align} uv-\int v\,du&=\frac12e^{2x}\sin x-\int\frac12e^{2x}\cos x\,dx\\ &=\frac12e^{2x}\sin x-\frac12\int e^{2x}\cos x\,dx\\ \end{align}$$

Let's use by parts again. LIATE tells us that $u$ should be the trig portion. $$\begin{array}{cc} \color{red}{u=\cos x}&v=\frac12e^{2x}\\ \downarrow&\uparrow\\ du=-\sin x\,dx&\color{red}{dv=e^{2x}\,dx} \end{array}$$

$$\begin{align} &\frac12e^{2x}\sin x-\frac12\int e^{2x}\cos x\,dx\\ &=\frac12e^{2x}\sin x-\frac12\left[uv-\int v\,du\right]\\ &=\frac12e^{2x}\sin x-\frac12\left[\frac12e^{2x}\cos x-\int \frac12e^{2x}\cdot-\sin x\,dx\right]\\ &=\frac12e^{2x}\sin x-\frac12\left[\frac12e^{2x}\cos x+\frac12\color{red}{\int e^{2x}\sin x\,dx}\right] \end{align}$$

Notice that this integral is the same one we started out solving for. By using integration by parts we looped back around to our original integral. So let's write this all out and simplify. $$\begin{align} \small{\color{red}{\int e^{2x}\sin x\,dx}}&=\small{\frac12e^{2x}\sin x-\frac12\left[\frac12e^{2x}\cos x+\frac12\color{red}{\int e^{2x}\sin x\,dx}\right]}\\ \small{\color{red}{\int e^{2x}\sin x\,dx}}&=\small{\frac12e^{2x}\sin x-\frac14e^{2x}\cos x-\frac14\color{red}{\int e^{2x}\sin x\,dx}}\\ \small{\color{red}{\int e^{2x}\sin x\,dx}+\frac14\color{red}{\int e^{2x}\sin x\,dx}}&=\small{\frac12e^{2x}\sin x-\frac14e^{2x}\cos x}\\ \small{\frac54\color{red}{\int e^{2x}\sin x\,dx}}&=\small{\frac12e^{2x}\sin x-\frac14e^{2x}\cos x}\\ \small{\color{red}{\int e^{2x}\sin x\,dx}}&=\small{\frac45\left[\frac12e^{2x}\sin x-\frac14e^{2x}\cos x\right]}\\ \small{\color{red}{\int e^{2x}\sin x\,dx}}&=\boxed{\small{\frac25e^{2x}\sin x-\frac15e^{2x}\cos x+C}}\\ \end{align}$$

We first look at the integral $\int x \cos{x}\ dx$ and figure out what $u$ and $dv$ are. When we figure out what $u$ is, we find $du$ (the derivative of $u$). When we figure out what $dv$ is, we find $v$ (the anti-derivative of $dv$). After we have $u$, $du$, $v$, and $dv$, we plug it into the formula and simplify. After we have our new equation, we have a new $u$ and $dv$. We find the derivative of the new $u$ to get $du$ and the new $v$ by taking the anti-derivative of the new $dv$. After we have the new $u$, $du$, $v$, and $dv$, we plug it into the formula.

(Video) Using Integration by Parts Twice

by PatrickJMT

Introduction

Here's a quick video by PatrickJMT where he solves an integration by parts problem that requires you to use the technique twice.

The integration by parts method tells us that: $$\int u\,dv=uv-\int v\,du$$

These problems aren't much harder but they can be tricky. So be careful when solving and write everything out.

Example

The problem that Patrick solves is: $$\int(\ln x)^2\,dx$$

Let's start by selecting our $u$ and $dv$. $$\begin{array}{cc} \color{red}{u=(\ln x)^2}&\\ \downarrow&\uparrow\\ &\color{red}{dv=dx} \end{array}$$

Now find $du$ and $v$. $$\begin{array}{cc} \color{red}{u=(\ln x)^2}&v=x\\ \downarrow&\uparrow\\ du=\frac{2\ln x}{x}\,dx=&\color{red}{dv=dx} \end{array}$$

Let's put them into the expression. $$\begin{align} uv-\int v\,du&=x(\ln x)^2-\int x\cdot\frac{2\ln x}{x}\,dx\\ &=x(\ln x)^2-2\int\ln x\,dx \end{align}$$

Now we need to use by parts again. Let's use the variables $s$ and $t$ for this one. $$\begin{array}{cc} \color{red}{s=\ln x}&t=x\\ \downarrow&\uparrow\\ ds=\frac{1}{x}\,dx=&\color{red}{dt=dx} \end{array}$$

Let's add these in and integrate. $$\begin{align} x(\ln x)^2-2\int\ln x\,dx&=x(\ln x)^2-2\left[st-\int t\,ds\right]\\ &=x(\ln x)^2-2\left[x\ln x-\int x\cdot\frac1x\,dx\right]\\ &=x(\ln x)^2-2\left[x\ln x-\int dx\right]\\ &=x(\ln x)^2-2\left[x\ln x-x\right]\\ &=\boxed{x(\ln x)^2-2x\ln x+2x+C} \end{align}$$