Tag: Integration by Parts

Establish the integration formula for arccot x

by RoRi

November 29, 2015

Establish that the following integration formula is correct:

$\int \operatorname{arccot} x \, dx = x \operatorname{arccot} x + \frac{1}{2} \log (1+x^2) + C.$

Proof. We can establish this formula using integration by parts. Let

$\begin{align*} u &= \operatorname{arccot} x & du &= -\frac{1}{1+x^2} \\ dv &= dx & v &= x \end{align*}$

where we established the formula for the derivative of $\operatorname{arccot}$ in this exercise (Section 6.22, Exercise #3). Then we have

$\begin{align*} \int \operatorname{arccot} x \, dx &= x \operatorname{arccot} x + \int \frac{x}{1+x^2} \, dx \\ &= x \operatorname{arccot} x + \frac{1}{2} \log (1+x^2) + C. \qquad \blacksquare \end{align*}$

Prove integration formulas for e^axcos (bx) and e^axsin (bx)

by RoRi

November 13, 2015

Let $a$ and $b$ be constants with at least one of them nonzero and define

$A = \int e^{ax} \cos (bx) \, dx, \qquad B = \int e^{ax} \sin (bx) \, dx.$

Using integration by parts, establish the following formulas for constants $C_1, C_2$ ,

$aA - bB = e^{ax} \cos (bx) + C_1, \qquad aB + bA = e^{ax} \sin (bx) + C_2.$

Using these formulas prove the following integration formulas,

$\begin{align*} \int e^{ax} \cos (bx) \, dx &= \frac{e^{ax} (a \cos (bx) + b \sin (bx))}{a^2 + b^2} + C, \\ \int e^{ax} \sin (bx) \, dx &= \frac{e^{ax} (a \sin (bx) - b \cos (bx))}{a^2 + b^2} + C. \end{align*}$

To establish the formula $aA - bB = e^{ax} \cos (bx) + C_1$ we use integration by parts letting

$\begin{align*} u &= \cos (bx), & du &= -b \sin (bx) \, dx \\ dv &= e^{ax} \, dx & v &= \frac{1}{a} e^{ax}. \end{align*}$

Then we can evaluate $A$ using the formula for integration by parts,

$\begin{align*} &&A &= \int e^{ax} \cos (bx) \, dx \\[9pt] \implies && A&= \frac{1}{a}e^{ax} \cos (bx) + \frac{b}{a} \int e^{ax} \sin (bx) \, dx + C_1 \\[9pt] \implies && aA &= e^{ax} \cos (bx) + bB + C_1 \\[9pt] \implies && aA - bB &= e^{ax} \cos (bx) + C_1. \end{align*}$

To establish the second formula, $aB + bA = e^{ax} \sin (bx) + C_2$ , we use integration by parts again. Let

$\begin{align*} u &= \sin (bx) & du &= b \cos (bx) \, dx \\ dv &= e^{ax} \, dx & v &= \frac{1}{a} e^{ax}. \end{align*}$

Then we have

$\begin{align*} && B&= \int e^{ax} \sin (bx) \, dx \\[9pt] \implies && B &= \frac{1}{a} e^{ax} \sin (bx) - \frac{b}{a} \int e^{ax} \cos (bx) \, dx + C_2 \\[9pt] \implies && aB &= e^{ax} \sin (bx) - bA + C_2 \\[9pt] \implies && aB + bA &= e^{ax} \sin (bx) + C_2. \end{align*}$

This establishes the two requested equations, now we prove the two integral identities.

Proof. Solving for $B$ in the second equation above we have

$aB + bA = e^{ax} \sin (bx) + C_2 \quad \implies \quad B = \frac{1}{a} \left( e^{ax} \sin (bx) - bA) + C_2.$

Plugging this into the first equation we have

$\begin{align*} && aA - bB &= e^{ax} \cos (bx) + C_1 \\[9pt] \implies && aA - \frac{b}{a} \left( e^{ax} \sin (bx) - bA + C_2 \right) &= e^{ax} \cos (bx) + C_1 \\[9pt] \implies && aA - \frac{b}{a} e^{ax} \sin (bx) + \frac{b^2}{a} A &= e^{ax} \cos (bx) + C \\[9pt] \implies && A \left( a + \frac{b^2}{a}\right) &= e^{ax} \cos (bx) + \frac{b}{a} e^{ax} \sin (bx) + C \\[9pt] \implies && A (a^2 + b^2) &= a e^{ax} \cos (bx) + b e^{ax} \sin (bx) + C \\[9pt] \implies && A &= \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} + C \\[9pt] \implies && \int e^{ax} \cos (bx) \, dx &= \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} + C. \end{align*}$

Next, for the second integral equation we are asked to prove, we use the formula we obtained for $B$ above,

$B = \frac{1}{a} \left( e^{ax} \sin (bx) - bA \right) + C.$

Then, we use the expression we obtained for $A$ into this,

$B = \frac{1}{a} \left( e^{ax} \sin (bx) - b \left( \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} \right) \right) + C.$

This implies,

$\begin{align*} && B &= \frac{1}{a} e^{ax} \sin (bx) - \frac{be^{ax} \cos (bx) + \frac{b^2}{a} e^{ax} \sin (bx)}{a^2+b^2} + C \\[9pt] \implies && B &= \frac{(a^2+b^2)e^{ax} \sin (bx) - (ab)e^{ax} \cos (bx) - b^2 e^{ax} \sin (bx)}{a(a^2+b^2)} + C \\[9pt] \implies && B &= \frac{a^2 e^{ax} \sin (bx) - (ab) e^{ax} \cos (bx)}{a(a^2+b^2)} + C \\[9pt] \implies && B &= \frac{a e^{ax} \sin (bx) - b e^{ax} \cos (bx)}{a^2+b^2} + C \\[9pt] \implies && \int e^{ax} \sin (bx) \, dx &= \frac{a e^{ax} \sin (bx) - b e^{ax} \cos (bx)}{a^2+b^2} + C. \qquad \blacksquare \end{align*}$

Evaluate the indefinite integral of e^{x^1/2}

by RoRi

November 11, 2015

Compute the following indefinite integral

$\int e^{\sqrt{x}} \, dx.$

To evaluate this integral we want to make a substitution. First multiply the numerator and denominator by $2 \sqrt{x}$ to obtain,

$\int e^{\sqrt{x}} \, dx = \int \frac{2\sqrt{x} e^{\sqrt{x}}}{2 \sqrt{x}} \, dx.$

Now define the function $u$ by

$u = e^{\sqrt{x}}, \qquad du &= \frac{e^{\sqrt{x}}}{2\sqrt{x}}.$

This implies

$\log u = \sqrt{x}.$

Therefore, using the method of substitution, we have

$\begin{align*} \int e^{\sqrt{x}} \, dx &= \int 2 \log u \, du \\ &= 2 (u \log u - u) + C \\ &= 2 \left( e^{\sqrt{x}} \log (e^{\sqrt{x}}) - e^{\sqrt{x}} \right) + C\\ &= 2 (\sqrt{x} - 1)e^{\sqrt{x}} + C. \end{align*}$

Evaluate the indefinite integral of x²e^-2x

by RoRi

November 11, 2015

Compute the following indefinite integral,

$\int x^2 e^{-2x} \, dx.$

To compute this integral we will integrate by parts twice. First, let

$\begin{align*} u &= x^2 & du &= 2x \, dx \\ dv &= e^{-2x} \, dx & v &= -\frac{1}{2} e^{-2x}. \end{align*}$

Therefore we have

$\int x^2 e^{-2x} \, dx = -\frac{x^2}{2} e^{-2x} + \int xe^{-2x} \, dx.$

To evaluate this next integral we use integration by parts a second time with

$\begin{align*} u &= x & du &= dx \\ dv &= e^{-2x}\, dx & v &= -\frac{1}{2}e^{-2x}. \end{align*}$

Giving us

$\begin{align*} \int xe^{-2x} \, dx &= -\frac{x}{2}e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \\ &= -\frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x} + C. \end{align*}$

So, putting this back into our formula we have

$\begin{align*} \int x^2 e^{-2x} \, dx &= -\frac{x^2}{2} e^{-2x} + \int xe^{-2x} \, dx \\ &= -\frac{x^2}{2} e^{-2x} - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x} + C \\ &= -\frac{1}{2} e^{-2x} \left( x^2 + x + \frac{1}{2} \right) + C. \end{align*}$

Evaluate the indefinite integral of x²e^x

by RoRi

November 11, 2015

Compute the following indefinite integral,

$\int x^2 e^x \, dx.$

We compute the integral using integration by parts. Let,

$\begin{align*} u &= x^2 & du &= 2x \, dx \\ dv &= e^x \, dx & v &= e^x. \end{align*}$

Then we have

$\int x^2 e^x \, dx &= x^2 e^x - 2 \int xe^x \, dx.$

But we know from a previous exercise (Section 6.17, Exercise #13) that

$\int xe^x \, dx = e^x (x-1) + C.$

Therefore we have,

$\begin{align*} \int x^2 e^x \, dx &= x^2 e^x - 2 \int xe^x \, dx \\ &= x^2 e^x - 2 \left( e^x(x-1) \right) + C \\ &= e^x (x^2 - 2x + 2) + C. \end{align*}$

Evaluate the indefinite integral of xe^-x

by RoRi

November 11, 2015

Compute the following indefinite integral,

$\int xe^{-x} \, dx.$

To evaluate this integral we use integration by parts, letting

$\begin{align*} u &= x & du &= dx, \\ dv &= e^{-x} \, dx & v &= -e^{-x}. \end{align*}$

Therefore,

$\begin{align*} \int xe^{-x} \, dx &= -xe^{-x} + \int e^{-x} \, dx \\ &= -xe^{-x} - e^{-x} + C \\ &= -e^{-x} (x + 1) + C. \end{align*}$

Evaluate the indefinite integral of xe^x

by RoRi

November 10, 2015

Compute the following indefinite integral:

$\int xe^x \, dx.$

To evaluate this integral we use integration by parts, letting

$\begin{align*} u &= x, & du &= dx,\\ dv &= e^x \, dx, & dv &= e^x. \end{align*}$

Then we have,

$\begin{align*} \int xe^x , dx &= xe^x - \int e^x \, dx \\ &= xe^x - e^x + C \\ &= e^x (x-1) + C. \end{align*}$

Prove a recursion formula for the integral of x^m logⁿ x

by RoRi

November 5, 2015

Prove the following recursion formula holds:

$\int x^m \log^n x \, dx = \frac{x^{m+1} \log^n x}{m+1} - \frac{n}{m+1} \int x^m \log^{n-1} x \, dx.$

Apply this formula to find the solution of

$\int x^3 \log^3 x \, dx.$

Proof. To obtain this recursion formula we proceed by integrating by parts. To that end, let

$\begin{align*} u &= \log^n x & du &= \frac{n \log^{n-1} x}{x} \, dx \\ dv &= x^m \, dx & v &= \frac{x^{m+1}}{m+1}. \end{align*}$

Then we have

$\begin{align*} \int x^m \log^n x \, dx &= \int u \, dv \\ &= uv - \int v \, du \\ &= \frac{x^{m+1}}{m+1} \log^n x - \frac{n}{m+1} \int x^{m+1} \frac{\log^{n-1} x}{x} \, dx \\ &= \frac{x^{m+1}}{m+1} \log^n x - \frac{n}{m+1} \int x^m \log^{n-1} x \, dx. \qquad \blacksquare \end{align*}$

Now, to find the solution of

$\int x^3 \log^3 x \, dx$

we apply the formula with $n = m = 3$ . This gives us,

$\int x^3 \log^3 x \, dx = \frac{x^4}{4} \log^3 |x| - \frac{3}{4} \int x^3 \log^2 x \, dx.$

Next we apply the formula to the resulting integral with $m = 3$ and $n = 2$ . Therefore we have

$\begin{align*} \int x^3 \log^3 x \, dx &= \frac{x^4}{4} \log^3 |x| - \frac{3}{4} \int x^3 \log^2 x \, dx \\ &= \frac{x^4}{4} \log^3 |x| - \frac{3}{4} \left( \frac{x^4}{4} \log^2 |x| - \frac{1}{2} \int x^3 \log x \, dx \right) \\ &= \frac{x^4}{4} \log^3 |x| - \frac{3x^4}{16} \log^2 |x| + \frac{3}{8} \int x^3 \log x \, dx. \end{align*}$

Finally, we apply the formula to the integral above with $m = 3$ and $n = 1$ to obtain,

$\begin{align*} \int x^3 \log^3 x \, dx &= \frac{x^4}{4} \log^3 |x| - \frac{3x^4}{16} \log^2 |x| + \frac{3}{8} \int x^3 \log x \, dx \\ &= \frac{x^4}{4} \log^3 |x| - \frac{3x^4}{16} \log^2 |x| + \frac{3}{8} \left( \frac{x^4}{4} \log |x| - \frac{1}{4} \int x^3 \, dx \right) \\ &= \frac{x^4}{4} \log^3 |x| - \frac{3x^4}{16} \log^2 |x| + \frac{3x^4}{32} \log |x| - \frac{3x^4}{128} + C. \end{align*}$

Find the integral of xⁿ log (ax)

by RoRi

November 4, 2015

Evaluate the following integral:

$\int x^n \log (ax) \, dx.$

Here we want to use integration by parts. To that end we define,

$\begin{align*} u &= \log (ax) & du &= \frac{1}{x} \, dx \\ dv &= x^n \, dx & v &= \frac{x^{n+1}}{n+1}, \end{align*}$

if $n \neq -1$ .

Therefore, integrating by parts we obtain the integral for all $n \neq -1$ .

$\begin{align*} \int x^n \log (ax) \, dx &= \int u \, dv \\ &= uv - \int v \, du \\ &= \frac{x^{n+1}}{n+1} \log |ax| - \frac{1}{n+1} \int x^n \, dx \\ &= \frac{x^{n+1}}{n+1} \log |ax| - \frac{x^{n+1}}{(n+1)^2} + C \end{align*}$

Now, for the case that $n = -1$ we have

$\int x^n \log (ax) \, dx = \int \frac{\log (ax)}{x} \, dx.$

Since $\frac{1}{x} \, dx$ is the derivative of $\log (ax)$ we make the substitution $u = \log (ax)$ and $du = \frac{dx}{x}$ to obtain

$\begin{align*} \int \frac{\log(ax)}{x} \, dx &= \int u \, du \\ &= \frac{1}{2} u^2 + C \\ &= \frac{1}{2} (\log |ax|)^2 + C. \end{align*}$

Find the integral of x log²x

by RoRi

November 3, 2015

Evaluate the following integral:

$\int x \log^2 x \, dx.$

To evaluate this integral we use integration by parts, defining

$\begin{align*} u &= \log^2 x & du &= \frac{2 \log x}{x} \, dx \\ dv &= x \, dx & v &= \frac{1}{2}x^2. \end{align*}$

Then we have

$\begin{align*} \int x \log^2 x \, dx &= \int u \, dv \\ &= uv - \int v \, du \\ &= \frac{1}{2}x^2 \log^2 |x| - \int x \log x \, dx. \end{align*}$

From the previous exercise (Section 6.9, #18) we know

$\int x \log x \, dx = \frac{1}{x^2} \log |x| - \frac{1}{4} x^2 + C.$

Therefore,

$\begin{align*} \int x \log^2 x \, dx &= \frac{1}{2}x^2 \log^2 |x| - \int x \log x \, dx \\ &= \frac{1}{2} x^2 \log^2 |x| - \frac{1}{2} x^2 \log |x| + \frac{1}{4}x^2 + C. \end{align*}$