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Anti-derivatives of Products - Integration by parts

The Product rule says:

$$\frac{d}{dx}\left(uv\right) = u\frac{dv}{dx}+v\frac{du}{dx}$$

Now, Lets take the anti derivative of both sides with respect to $dx$.

$$\int\frac{d}{dx}\left(uv\right)dx = \int u\frac{dv}{dx}+v\frac{du}{dx} dx$$ $$\int\frac{d}{dx}\left(uv\right)dx = \int u\frac{dv}{dx}dx+\int v\frac{du}{dx} dx$$ $$uv = \int u\frac{dv}{dx}dx+\int v\frac{du}{dx} dx$$

Rearranging:

$$\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx} dx$$

Lets try using the rule above!

Find:

$$\int x\cos x \,dx$$

Let $x=u$ and $\cos x = \frac{dv}{dx}$. So we have:

$$\begin{aligned} \int x\cos x \,dx &= x\sin x - \int \sin x \,dx \\ &= x\sin x - (-\cos x) \\ &= x\sin x +\cos x \end{aligned}$$

NOT FINISHED: WILL BE FURTHER EDITED SUBSEQUENTLY.

It can be very difficult, or sometimes even impossible, to integrate a product. For example, $\int x\sin x dx$ has no elementary antiderivative.

There are two general methods to integrate $\int f(x)g(x) dx$ for two functions $f$ and $g$. Method 2 deals with various substitutions one can make to simplify an integral, which are too complicated to be displayed here.

Method 1: Integration by parts.

If either of $f$ or $g$ has an easy antiderivative (e.g. $\sin x$, $\cos x$, $x^2$), and the other function has an easy derivative (e.g. $arctan x$, $x^{10}$, $\ln x$), then consider integrating as in the following formula:

$\int u dv$ = $uv - \int v du$.

Example 1: Integrate $f(x) = \ln x$.

Solution 1: $x$ has an easy antiderivative, and $\ln x$ has an easy derivative.

$\int \ln x dx = x \ln x - \int x d(\ln x) = x \ln x - \int x * \frac{1}{x} dx = x \ln x - x + C$.

Exercise 1. Integrate $f(x) = \arctan x$. (Hint: Use integration by parts to transform the integral into one you know how to solve.)

Example 2: Integrate $f(x) = x^2\sin x$.

Solution 2: This example will demonstrate that multiple integrations by parts can successfully integrate an function.

$\int x^2\sin x dx = \int -x^2d(\cos x) \\= -x^2\cos x + \int \cos x d(x^2) = -x^2\cos x + \int 2x \cos x dx.$

$\int x \cos x dx = \int x d(\sin x) = x \sin x - \int \sin x dx = x \sin x + \cos x + C$.

Substituting the result of the second integral into the first expression gives

$\int x^2\sin x dx = -x^2\cos x + 2x \sin x + 2\cos x + C.$

Exercise 2. Verify that the derivative of $-x^2\cos x + 2x \sin x + 2\cos x + C$ equals $x^2\sin x.$

Exercise 3. Integrate the function $f(x) = e^x \sin x$ using integration by parts. (Hint: Set the integral expression equal to an arbitrary variable $I$.)

Exercise 4. Integrate the function $f(x) = x^{2014} e^{2x}$, using sigma (sum) notation if necessary.

Exercise 5. Integrate the function $f(x) = x^2 \ln x$. Generalize to $f(x) = x^n \ln x$.

Exercise 6. Find the area of the region in the interval $[-1, 1]$ of $f(x) = \arcsin x$, using the convention that areas under the x-axis have negative area.

Here is some practice for integration by parts. Best of luck!