Integration by Parts (Solutions to Problems)

The following solutions are for the integration practice problems linked here. I encourage you to attempt them by yourself first before looking through the solutions.

Original problems statements are highlighted in blue. Final answers are highlighted in green.

Problem 1

Evaluate the following integral expressions. Note: For some of these problems, you may have to perform integration by parts successively multiple times.

1. $\int_1^9 \frac{\ln x}{\sqrt{x}}dx$

Define $u=\ln x$ and $v=2\sqrt{x}$, such that

\[\frac{du}{dx}=\frac{1}{x} \longrightarrow du=\frac{1}{x}dx, \quad \frac{dv}{dx}=\frac{2}{2\sqrt{x}}=\frac{1}{\sqrt{x}} \longrightarrow dv=\frac{1}{\sqrt{x}}dx\]

Using these function definitions, we can conclude using integration by parts that

\[\int_1^9\frac{\ln x}{\sqrt{x}}dx=\int_1^9 udv=uv\vert_1^9-\int_1^9 vdu=2\ln(x)\sqrt{x}\vert_1^9-\int_1^9\frac{2\sqrt{x}}{x}dx\]

Simplifying the right hand side, we find

\[\int_1^9\frac{\ln x}{\sqrt{x}}dx=6\ln(9)-2\ln(1)-2\int_1^9x^{-1/2}dx=6\ln(9)-2\int_1^9x^{-1/2}dx\]

Using the power rule for integration, we can simplify the remaining integral on the right hand side.

\[\int_1^9\frac{\ln x}{\sqrt{x}}dx=6\ln(9)-4\sqrt{x}\vert_1^9=6\ln(9)-12+4=-8+6\ln9\]

Therefore, we can conclude that

$$\int_1^9 \frac{\ln x}{\sqrt{x}}dx=-8+6\ln 9$$

2. $\int x^2\sin(x)dx$

Choose $u=x^2$ and $v=-\cos x$, such that

\[\frac{du}{dx}=2x\longrightarrow du=2xdx, \quad \frac{dv}{dx}=\sin x \longrightarrow dv=\sin xdx\]

Using these function definitions, we can conclude using integration by parts that

\[\int x^2\sin(x)dx=\int udv= uv-\int vdu=-x^2\cos x+2\int x\cos xdx\]

As per the hint given in the problem, let’s try doing integration by parts one more on the integral on the right hand side, now defining $u=x$ and $v=\sin x$, such that

\[\frac{du}{dx}=1\longrightarrow du=dx, \quad \frac{dv}{dx}=\cos x \longrightarrow dv=\cos x dx\]

Using these function definitions, we can conclude using integration by parts that

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

Now, we can easily evaluate the remaining integral on the right hand side to give

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

3. $\int \sin^{-1} (x)dx$

Using the hint given, we can choose $u=\sin^{-1}(x)$ and $v=x$, such that

\[\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}} \longrightarrow du=\frac{dx}{\sqrt{1-x^2}}, \quad \frac{dv}{dx}=1 \longrightarrow dv=dx\]

Using these function definitions, we can conclude using integration by parts that

\[\int \sin^{-1} (x)dx=\int udv=uv-\int vdu=x\sin^{-1}(x)-\int \frac{xdx}{\sqrt{1-x^2}}\]

To evaluate the remaining integral on the right hand side, we can use $u$-substitution. Define $u=1-x^2$ such that $\frac{du}{dx}=-2x$ and hence $-\frac{1}{2}du=xdx$. Making this substitution above, we have

\[\int \sin^{-1} (x)dx=x\sin^{-1}(x)-\int -\frac{1}{2}\frac{du}{\sqrt{u}}=x\sin^{-1}(x)+\frac{1}{2}\int u^{-1/2}du\]

Using the power rule for integration, we have

\[\int \sin^{-1} (x)dx=x\sin^{-1}(x)+\sqrt{u}+C\]

Substituting back our definition for $u=1-x^2$ from above, we have

$$\int \sin^{-1} (x)dx=x\sin^{-1}(x)+\sqrt{1-x^2}+C$$

Problem 2

Evaluate the following integral expression.

$$\int_0^\pi e^x\cos xdx$$

Let us define $u=e^x$ and $v=\sin x$, such that $du=e^xdx$ and $dv=\cos xdx$. Therefore, we can rewrite the given integral using integration by parts

\[\int_0^\pi e^x\cos xdx=e^x\sin x\vert_0^\pi-\int_0^{\pi} e^x\sin xdx\]

Note that $e^x\sin x\vert_0^\pi=e^{\pi}\sin\pi-e^0\sin 0=0-0=0$, since $\sin 0=\sin \pi=0$. Therefore, the right hand side of this equation simplifies to

\[\int_0^\pi e^x\cos xdx=-\int_0^{\pi} e^x\sin xdx\]

We can do integration by parts again on the integral expression on the right hand side, defining $u=e^x$ and $v=\cos x$ such that $du=e^xdx$ and $dv=-\sin xdx$. Therefore, we can use integration by parts once more to find

\[\int_0^\pi e^x\cos xdx=-\int_0^{\pi} e^x\sin xdx=e^x\cos x\vert_0^\pi-\int_0^{\pi} e^x\cos x dx\]

Note that we can add $\int_0^{\pi} e^x\cos x dx$ to both sides of this equation to find

\[2\int_0^{\pi} e^x\cos x dx=e^x\cos x\vert_0^\pi=e^{\pi}\cos(\pi)-e^0\cos0=-e^\pi-1=-(1+e^{\pi})\]

Dividing both sides by $2$, we find

$$\int_0^{\pi} e^x\cos x dx=\frac{-(1+e^{\pi})}{2}$$

Our method of obtaining this solution is interesting to comment on. Notice that in this entire process, we never technically evaluated any integral! All we did was use integral by parts twice to show that $I=(\text{something})-I$, where $I$ is the original integral expression, and then solve for $I$. This is a common trick that can be used in conjunction with integration by parts: rewrite the initial integral as a closed-form function of itself and then solve for the initial integral.

Problem 3

If $n$ is a positive integer, prove that

$$\int_0^1 (\ln x)^ndx=(-1)^n n!$$

This problem looks somewhat familiar to the practice problem from earlier on evaluating the integral expression $\int \ln(x) dx$, so we may take inspiration from that problem first in initially choosing our $u$ and $v$ functions. Let’s first guess that $u=(\ln x)^n$ and $v=x$, such that

\[\frac{du}{dx}=n(\ln x)^{n-1}\frac{1}{x} \longrightarrow du=\frac{n(\ln x)^{n-1}}{x}dx, \quad \frac{dv}{dx}=1 \longrightarrow dv=dx\]

Using these definitions of $u$ and $v$, we can rewrite the integral and use integration by parts:

\[\int_0^1 (\ln x)^ndx=\int_0^1 udv=\left.uv\right\vert_0^1-\int_0^1 vdu=\left.x(\ln x)^n\right\vert_0^1-\int_0^1dx\frac{n(\ln x)^{n-1}}{x}x\]

Let’s break down the right hand side, and first separately consider the first expression $\left.x(\ln x)^n\right\vert_0^1$:

\[\left.x(\ln x)^n\right\vert_0^1=(\ln 1)^n-\lim_{x\rightarrow 0} x(\ln x)^n\]

$\ln 1=0$ and $0^n=0$ for any positive integer $n$. The remaining limit is not as straightforward to quantify, although we can use L’Hopital’s Rule to rewrite it:

\[\lim_{x\rightarrow 0} x(\ln x)^n=\lim_{x\rightarrow 0} \frac{(\ln x)^n}{\frac{1}{x}}\]

Direct substitution gives the indefinite form $-\infty/\infty$, so we can use L’Hopital’s Rule to get

\[\lim_{x\rightarrow 0} x(\ln x)^n=\lim_{x\rightarrow 0}\frac{\frac{n(\ln x)^{n-1}}{x}}{\frac{-1}{x^2}}=-n\lim_{x\rightarrow 0}x(\ln x)^{n-1}\]

Notice that aside from the prefactor of $-n$, this limit looks almost identical to the original limit $\lim_{x\rightarrow 0} x(\ln x)^n$ that we were trying to evaluate, just with the power of $\ln x$ decreased by one. This allows us to guess that perhaps we can used L’Hopital’s Rule again, and once more, and so one and so forth:

$$\lim_{x\rightarrow 0} x(\ln x)^n=-n\lim_{x\rightarrow 0}x(\ln x)^{n-1}=n(n-1)\lim_{x\rightarrow 0}x(\ln x)^{n-2}=-n(n-1)(n-2)\lim_{x\rightarrow 0}x(\ln x)^{n-3}=\cdots$$

Ultimately, we can begin to see pattern emerge: the signs alternate between negative and positive, and we ultimately accumulate the prefactor $n(n-1)(n-2)\cdots 2\cdot 1=n!$. Therefore, we find

\[\lim_{x\rightarrow 0} x(\ln x)^n=(-1)^nn!\lim_{x\rightarrow 0}x(\ln x)^0=(-1)^nn!\lim_{x\rightarrow 0}x=(-1)^nn!\cdot 0=0\]

since $(\ln x)^0=1$ for any positive $x$. All in all, we have shown that this limit expression is equal to $0$, and so returning to our integration be parts problem, we have shown that

\[\left.x(\ln x)^n\right\vert_0^1=(\ln 1)^n-\lim_{x\rightarrow 0} x(\ln x)^n=0-0=0\]

and so

\[\int_0^1 (\ln x)^ndx=\left.x(\ln x)^n\right\vert_0^1-\int_0^1dx\frac{n(\ln x)^{n-1}}{x}x=-\int_0^1dx\frac{n(\ln x)^{n-1}}{x}x\]

Now, let’s simplify the remaining integral expression on the right hand side:

\[\int_0^1 (\ln x)^ndx=-n\int_0^1dx(\ln x)^{n-1}\]

Let us denote the integral expression $I_n=\int_0^1 (\ln x)^ndx$. This equation has effectively shown that $I_n=-nI_{n-1}$. This also means that $I_{n-1}=-(n-1)I_{n-2}$, and $I_{n-2}=-(n-2)I_{n-3}$, and so on and so forth. We can use this recursive definition to find the following:


This pattern continues all the way until we reach $n=0$:


What is $I_0$? Citing our definition for $I_n$, we know that $I_0=\int_0^1 (\ln x)^0 dx$, but since any number to the zero power is $1$, this simplifies to

\[I_0=\int_0^1 dx=\left. x\right\vert_0^1=1-0=1\]

Therefore, we can conclude that $I_n=(-1)^nn!I_0=(-1)^nn!(1)=(-1)^nn!$, and so

If $n$ is a positive integer, then

$$\int_0^1 (\ln x)^ndx=(-1)^n n!$$

Problem 4

Show that


Hint: Start by showing that if $I_n$ denotes the integral, then


We can try $u=(1-x^2)^n$ and $v=x$, such that

$$\frac{du}{dx}=-2xn(1-x^2)^{n-1}\longrightarrow du=-2nx(1-x^2)^{n-1}dx, \quad \frac{dv}{dx}=1\longrightarrow dv=dx$$

This allows us to rewrite the initial integral expression using integration by parts as

$$\int_0^1(1-x^2)^ndx=\int_0^1 udv=\left.uv\right\vert_0^1-\int_0^1 vdu=\left.x(1-x^2)^n\right\vert_0^1-\int_0^1 -2nx^2(1-x^2)^{n-1}dx$$

Let’s break down the right hand side, and first separately consider the first expression $\left.x(1-x^2)^n\right\vert_0^1$:


Therefore, we can conclude that

\[\int_0^1(1-x^2)^ndx=-\int_0^1 -2nx^2(1-x^2)^{n-1}dx=2n\int_0^1 x^2(1-x^2)^{n-1}dx\]

This next step is the “trick” to this problem: we’re going to subtract the expression $2n\int_0^1 (1-x^2)^{n-1}dx$ from both sides of the equation.

$$\int_0^1(1-x^2)^ndx-2n\int_0^1 (1-x^2)^{n-1}dx=-2n\int_0^1 (1-x^2)^{n-1}dx+2n\int_0^1 x^2(1-x^2)^{n-1}dx$$

Note that we can now combine the two integral expressions on the right hand side:

$$\int_0^1(1-x^2)^ndx-2n\int_0^1 (1-x^2)^{n-1}dx=-2n\int_0^1 (1-x^2)(1-x^2)^{n-1}dx=-2n\int_0^1 (1-x^2)^{n}dx$$

Rearranging both sides of this equation, we find that

\[(2n+1)\int_0^1(1-x^2)^ndx=2n\int_0^1 (1-x^2)^{n-1}dx\]

In other words,

\[\int_0^1(1-x^2)^ndx =\frac{2n}{2n+1}\int_0^1 (1-x^2)^{n-1}dx\]

Let us reindex this equation by defining a new index $k$ such that $n=k+1$. This gives us

\[\int_0^1(1-x^2)^{k+1}dx =\frac{2(k+1)}{2(k+1)+1}\int_0^1 (1-x^2)^{(k+1)-1}dx\]


\[\int_0^1(1-x^2)^{k+1}dx =\frac{2k+2}{2k+3}\int_0^1 (1-x^2)^{k}dx\]

If we define the function $I_k=\int_0^1 (1-x^2)^{k}dx$, the original expression for the integral we’re trying to evaluate with $n=k$, then

\[I_{k+1} =\frac{2k+2}{2k+3}I_k\]

This gives us the result as expected from the hint. Because this equation is true for general $k$, this means that we also know from this equation that \(I_{k} =\frac{2k}{2k+1}I_{k-1}\), \(I_{k-1} =\frac{2k-2}{2k-1}I_{k-2}\), and so on and so forth until we have $I_1=\frac{2}{3}I_0$. This recursive relation allows us to write


Note that the prefactor can be rewritten by multiplying both the numerator and denominator by $2k(2k-2)(2k-4)\cdots4\cdot2$.

\[I_k=\frac{(2k)^2(2k-2)^2(2k-4)^2\cdots 4^22^2}{(2k+1)(2k)(2k-1)(2k-2)(2k-3)\cdots5\cdot4\cdot3\cdot2}I_0\]

The denominator is now clearly $(2k+1)!$. The numerator requires a bit of additional work. Note that

\[I_k=\frac{\left[2k(2k-2)(2k-4)\cdots 4\cdot2\right]^2}{(2k+1)!}I_0\]

Since every factor in the numerator is even, let’s pull out a factor of $2$ from every single factor in the numerator. There are $k$ factors, and so

\[I_k=\frac{\left[2^kk(k-1)(k-2)\cdots 2\cdot1\right]^2}{(2k+1)!}I_0=\frac{\left[2^kk!\right]^2}{(2k+1)!}I_0\]

by the definition of $k!$. Rewriting, we have


Now, all we have to do is evaluate $I_0$. Using the definition of $I_k=\int_0^1 (1-x^2)^{k}dx$ from above, this means that

\[I_0=\int_0^1 (1-x^2)^0dx=\int_0^1 dx=1-0=1\]

since anything to the zero power is $1$. Therefore, we can conclude that


Renaming the variable from $k\rightarrow n$ gives us the desired result: