15 Integration by parts exercises solved step by step

Set \(u = x\) and \(dv = e^{3x}\,dx\). Then \(du = dx\) and \(v = \dfrac{e^{3x}}{3}\).

\[
\int x e^{3x}\,dx = \frac{x e^{3x}}{3} – \int \frac{e^{3x}}{3}\,dx
= \frac{x e^{3x}}{3} – \frac{e^{3x}}{9} + C.
\]
Solution to question 2:

Differentiate \(F(x) = \dfrac{x e^{3x}}{3} – \dfrac{e^{3x}}{9}\):

\[
F'(x) = \frac{e^{3x} + 3x e^{3x}}{3} – \frac{3e^{3x}}{9}
= \frac{e^{3x}}{3} + x e^{3x} – \frac{e^{3x}}{3} = x e^{3x}. \checkmark
\]

Let \(u = x\), \(dv = \cos(2x)\,dx\). Then \(du = dx\), \(v = \dfrac{\sin(2x)}{2}\).

\[
\int x\cos(2x)\,dx = \frac{x\sin(2x)}{2} – \int \frac{\sin(2x)}{2}\,dx
= \frac{x\sin(2x)}{2} + \frac{\cos(2x)}{4} + C.
\]
Solution to question 2:

Let \(u = 3x-1\), \(dv = \sin(x)\,dx\). Then \(du = 3\,dx\), \(v = -\cos(x)\).

\[
\int (3x-1)\sin(x)\,dx = -(3x-1)\cos(x) + 3\int \cos(x)\,dx
= -(3x-1)\cos(x) + 3\sin(x) + C.
\]

First pass: \(u = x^2\), \(dv = e^{-x}\,dx \Rightarrow du = 2x\,dx\), \(v = -e^{-x}\).

\[
\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int x e^{-x}\,dx.
\]

Second pass: \(u = x\), \(dv = e^{-x}\,dx \Rightarrow du = dx\), \(v = -e^{-x}\).

\[
\int x e^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} – e^{-x} + C.
\]

Combining:

\[
\int x^2 e^{-x}\,dx = -x^2 e^{-x} – 2xe^{-x} – 2e^{-x} + C
= -e^{-x}(x^2 + 2x + 2) + C.
\]

Solution to question 2:

Let \(u = x^2 + 2x\), \(dv = e^{2x}\,dx \Rightarrow du = (2x+2)\,dx\), \(v = \dfrac{e^{2x}}{2}\).

\[
\int (x^2+2x)e^{2x}\,dx = \frac{(x^2+2x)e^{2x}}{2} – \int \frac{(2x+2)e^{2x}}{2}\,dx.
\]

Now apply once more with \(u = x+1\), \(dv = e^{2x}\,dx\):

\[
\int (x+1)e^{2x}\,dx = \frac{(x+1)e^{2x}}{2} – \frac{e^{2x}}{4} + C.
\]

Combining:

\[
\int (x^2+2x)e^{2x}\,dx = \frac{(x^2+2x)e^{2x}}{2} – \frac{(x+1)e^{2x}}{2} + \frac{e^{2x}}{4} + C
= e^{2x}\!\left(\frac{x^2+x}{2} – \frac{1}{4}\right) + C.
\]

Let \(u = \ln x\), \(dv = dx \Rightarrow du = \tfrac{1}{x}\,dx\), \(v = x\).

\[
\int \ln x\,dx = x\ln x – \int x \cdot \frac{1}{x}\,dx = x\ln x – \int 1\,dx = x\ln x – x + C.
\]

Solution to question 2:

Using \(\ln(5x) = \ln 5 + \ln x\):

\[
\int \ln(5x)\,dx = \int \ln 5\,dx + \int \ln x\,dx = x\ln 5 + x\ln x – x + C = x\ln(5x) – x + C.
\]

Let \(u = \ln x\), \(dv = x^3\,dx \Rightarrow du = \tfrac{1}{x}\,dx\), \(v = \tfrac{x^4}{4}\).

\[
\int x^3 \ln x\,dx = \frac{x^4 \ln x}{4} – \int \frac{x^4}{4} \cdot \frac{1}{x}\,dx
= \frac{x^4 \ln x}{4} – \frac{1}{4}\int x^3\,dx
= \frac{x^4 \ln x}{4} – \frac{x^4}{16} + C.
\]

Solution to question 2:

Let \(u = \ln(x^2+1)\), \(dv = x\,dx \Rightarrow du = \tfrac{2x}{x^2+1}\,dx\), \(v = \tfrac{x^2}{2}\).

\[
\int x\,\ln(x^2+1)\,dx = \frac{x^2}{2}\ln(x^2+1) – \int \frac{x^3}{x^2+1}\,dx.
\]

Perform polynomial division: \(\dfrac{x^3}{x^2+1} = x – \dfrac{x}{x^2+1}\).

\[
\int \frac{x^3}{x^2+1}\,dx = \frac{x^2}{2} – \frac{1}{2}\ln(x^2+1) + C.
\]

Therefore:

\[
\int x\,\ln(x^2+1)\,dx = \frac{x^2}{2}\ln(x^2+1) – \frac{x^2}{2} + \frac{\ln(x^2+1)}{2} + C.
\]

Let \(u = \arctan x\), \(dv = dx \Rightarrow du = \tfrac{dx}{1+x^2}\), \(v = x\).

\[
\int \arctan x\,dx = x\arctan x – \int \frac{x}{1+x^2}\,dx
= x\arctan x – \frac{1}{2}\ln(1+x^2) + C.
\]

Solution to question 2:

Let \(u = \arcsin x\), \(dv = dx \Rightarrow du = \tfrac{dx}{\sqrt{1-x^2}}\), \(v = x\).

\[
\int \arcsin x\,dx = x\arcsin x – \int \frac{x}{\sqrt{1-x^2}}\,dx.
\]

Substitute \(t = 1 – x^2\), \(dt = -2x\,dx\):

\[
\int \frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2} + C.
\]

Therefore:

\[
\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C.
\]

Let \(u = \arctan x\), \(dv = x\,dx \Rightarrow du = \tfrac{dx}{1+x^2}\), \(v = \tfrac{x^2}{2}\).

\[
\int x\,\arctan x\,dx = \frac{x^2}{2}\arctan x – \frac{1}{2}\int \frac{x^2}{1+x^2}\,dx.
\]

Since \(\dfrac{x^2}{1+x^2} = 1 – \dfrac{1}{1+x^2}\):

\[
\int x\,\arctan x\,dx = \frac{x^2}{2}\arctan x – \frac{x}{2} + \frac{\arctan x}{2} + C
= \frac{x^2+1}{2}\arctan x – \frac{x}{2} + C.
\]

Solution to question 2:

Let \(u = \arcsin x\), \(dv = x^2\,dx \Rightarrow du = \tfrac{dx}{\sqrt{1-x^2}}\), \(v = \tfrac{x^3}{3}\).

\[
\int x^2 \arcsin x\,dx = \frac{x^3}{3}\arcsin x – \frac{1}{3}\int \frac{x^3}{\sqrt{1-x^2}}\,dx.
\]

For the residual integral, write \(x^3 = x^2 \cdot x\) and substitute \(t = 1 – x^2\), so \(x^2 = 1-t\) and \(x\,dx = -\tfrac{dt}{2}\):

\[
\int \frac{x^3}{\sqrt{1-x^2}}\,dx = \int \frac{(1-t)}{\sqrt{t}}\cdot\left(-\frac{1}{2}\right)dt
= \frac{1}{2}\int\!\left(t^{1/2} – t^{-1/2}\right)dt = \frac{(1-x^2)^{3/2}}{3} – \sqrt{1-x^2} + C.
\]

Combining:

\[
\int x^2 \arcsin x\,dx = \frac{x^3}{3}\arcsin x – \frac{(1-x^2)^{3/2}}{9} + \frac{\sqrt{1-x^2}}{3} + C.
\]

\[
\int x^3 e^x\,dx = e^x(x^3 – 3x^2 + 6x – 6) + C.
\]

Solution to question 2:

\[
\int x^2 \cos x\,dx = x^2 \sin x + 2x\cos x – 2\sin x + C.
\]

\[
\int x^4 \sin x\,dx = -x^4\cos x + 4x^3\sin x + 12x^2\cos x – 24x\sin x – 24\cos x + C.
\]

Solution to question 2:

Let \(p(x) = 2x^3 – x\). Build the table with \(dv = \cos(3x)\,dx\), so antiderivatives cycle through \(\tfrac{\sin(3x)}{3}, -\tfrac{\cos(3x)}{9}, -\tfrac{\sin(3x)}{27}, \tfrac{\cos(3x)}{81}\).

\[
\int (2x^3-x)\cos(3x)\,dx = \frac{(2x^3-x)\sin(3x)}{3} + \frac{(6x^2-1)\cos(3x)}{9}
– \frac{12x\sin(3x)}{27} – \frac{12\cos(3x)}{81} + C.
\]

Simplifying the last term: \(\dfrac{12}{81} = \dfrac{4}{27}\).

\[
= \frac{(2x^3-x)\sin(3x)}{3} + \frac{(6x^2-1)\cos(3x)}{9} – \frac{4x\sin(3x)}{9} – \frac{4\cos(3x)}{27} + C.
\]

Let \(I = \int e^x \sin x\,dx\). Set \(u = e^x\), \(dv = \sin x\,dx\):

\[
I = -e^x\cos x + \int e^x\cos x\,dx.
\]

Apply again with \(u = e^x\), \(dv = \cos x\,dx\):

\[
I = -e^x\cos x + e^x\sin x – \int e^x\sin x\,dx = -e^x\cos x + e^x\sin x – I.
\]

Solving:

\[
2I = e^x(\sin x – \cos x) \implies \boxed{I = \frac{e^x(\sin x – \cos x)}{2} + C}.
\]

Solution to question 2:

Let \(I = \int e^{2x}\cos(3x)\,dx\). Let \(u = e^{2x}\), \(dv = \cos(3x)\,dx\):

\[
I = \frac{e^{2x}\sin(3x)}{3} – \frac{2}{3}\int e^{2x}\sin(3x)\,dx.
\]

Apply again with \(u = e^{2x}\), \(dv = \sin(3x)\,dx\):

\[
\int e^{2x}\sin(3x)\,dx = -\frac{e^{2x}\cos(3x)}{3} + \frac{2}{3}\int e^{2x}\cos(3x)\,dx
= -\frac{e^{2x}\cos(3x)}{3} + \frac{2I}{3}.
\]

Substituting back:

\[
I = \frac{e^{2x}\sin(3x)}{3} – \frac{2}{3}\!\left(-\frac{e^{2x}\cos(3x)}{3} + \frac{2I}{3}\right)
= \frac{e^{2x}\sin(3x)}{3} + \frac{2e^{2x}\cos(3x)}{9} – \frac{4I}{9}.
\]
\[
\frac{13I}{9} = \frac{3e^{2x}\sin(3x) + 2e^{2x}\cos(3x)}{9}
\implies \boxed{I = \frac{e^{2x}(3\sin(3x) + 2\cos(3x))}{13} + C}.
\]

Let \(u = x\), \(dv = e^x\,dx \Rightarrow du = dx\), \(v = e^x\).

\[
\int_0^1 x e^x\,dx = \bigl[xe^x\bigr]_0^1 – \int_0^1 e^x\,dx
= (1\cdot e – 0) – \bigl[e^x\bigr]_0^1 = e – (e – 1) = 1.
\]

Solution to question 2:

Let \(u = x\), \(dv = \sin(x)\,dx \Rightarrow du = dx\), \(v = -\cos(x)\).

\[
\int_0^{\pi} x\sin(x)\,dx = \bigl[-x\cos(x)\bigr]_0^{\pi} + \int_0^{\pi}\cos(x)\,dx
= (-\pi\cos\pi + 0) + \bigl[\sin x\bigr]_0^{\pi}.
\]
\[
= \pi + (\sin\pi – \sin 0) = \pi + 0 = \pi.
\]

Let \(u = \ln x\), \(dv = x\,dx \Rightarrow du = \tfrac{dx}{x}\), \(v = \tfrac{x^2}{2}\).

\[
\int_1^e x\ln x\,dx = \left[\frac{x^2\ln x}{2}\right]_1^e – \int_1^e \frac{x}{2}\,dx
= \frac{e^2}{2} – 0 – \left[\frac{x^2}{4}\right]_1^e = \frac{e^2}{2} – \frac{e^2-1}{4}
= \frac{e^2+1}{4}.
\]

Solution to question 2:

Using the antiderivative from Exercise 4: \(\int \ln x\,dx = x\ln x – x\).

\[
\int_1^e \ln x\,dx = \bigl[x\ln x – x\bigr]_1^e = (e\cdot 1 – e) – (1\cdot 0 – 1) = 0 + 1 = 1.
\]

Let \(t = \sqrt{x}\), so \(x = t^2\) and \(dx = 2t\,dt\).

\[
\int \sqrt{x}\,e^{\sqrt{x}}\,dx = \int t\,e^t\cdot 2t\,dt = 2\int t^2 e^t\,dt.
\]

Using the tabular method on \(2\int t^2 e^t\,dt\):

\[
2\int t^2 e^t\,dt = 2e^t(t^2 – 2t + 2) + C.
\]

Substituting back \(t = \sqrt{x}\):

\[
\int \sqrt{x}\,e^{\sqrt{x}}\,dx = 2e^{\sqrt{x}}(x – 2\sqrt{x} + 2) + C.
\]

Solution to question 2:

Let \(t = \ln x\), so \(x = e^t\) and \(dx = e^t\,dt\).

\[
\int \sin(\ln x)\,dx = \int e^t\sin(t)\,dt.
\]

This is the cyclic integral from Exercise 10, question 1:

\[
\int e^t\sin(t)\,dt = \frac{e^t(\sin t – \cos t)}{2} + C.
\]

Substituting back \(t = \ln x\):

\[
\int \sin(\ln x)\,dx = \frac{x\bigl(\sin(\ln x) – \cos(\ln x)\bigr)}{2} + C.
\]

Let \(u = \sin^{n-1}x\), \(dv = \sin x\,dx \Rightarrow du = (n-1)\sin^{n-2}x\cos x\,dx\), \(v = -\cos x\).

\[
\int \sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2 x\,dx.
\]

Replace \(\cos^2 x = 1 – \sin^2 x\):

\[
= -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx – (n-1)\int \sin^n x\,dx.
\]

Move the last term to the left side:

\[
n\int \sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx,
\]
\[
\therefore \int \sin^n x\,dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x\,dx. \quad \checkmark
\]

Solution to question 2:

Apply with \(n = 4\) on \([0, \pi/2]\). At both endpoints \(\sin^{n-1}x\cos x\) vanishes, so the boundary term is always 0 on this interval.

\[
\int_0^{\pi/2}\sin^4 x\,dx = \frac{3}{4}\int_0^{\pi/2}\sin^2 x\,dx.
\]

Apply with \(n = 2\):

\[
\int_0^{\pi/2}\sin^2 x\,dx = \frac{1}{2}\int_0^{\pi/2}1\,dx = \frac{\pi}{4}.
\]

Therefore:

\[
\int_0^{\pi/2}\sin^4 x\,dx = \frac{3}{4}\cdot\frac{\pi}{4} = \frac{3\pi}{16}.
\]

Let \(u = x^2\), \(dv = x\sqrt{x^2+1}\,dx\). To find \(v\): set \(s = x^2+1\), \(ds = 2x\,dx\), so

\[
v = \int x\sqrt{x^2+1}\,dx = \frac{(x^2+1)^{3/2}}{3}.
\]

Also \(du = 2x\,dx\). Applying integration by parts:

\[
\int x^3\sqrt{x^2+1}\,dx = \frac{x^2(x^2+1)^{3/2}}{3} – \frac{2}{3}\int x(x^2+1)^{3/2}\,dx.
\]

For the remaining integral, use \(s = x^2+1\) again:

\[
\int x(x^2+1)^{3/2}\,dx = \frac{(x^2+1)^{5/2}}{5}.
\]

Therefore:

\[
\int x^3\sqrt{x^2+1}\,dx = \frac{x^2(x^2+1)^{3/2}}{3} – \frac{2(x^2+1)^{5/2}}{15} + C.
\]

Solution to question 2:

Let \(t = x^{1/3}\), so \(x = t^3\) and \(dx = 3t^2\,dt\).

\[
\int e^{\sqrt[3]{x}}\,dx = 3\int t^2 e^t\,dt.
\]

From Exercise 8, question 1 (with base function \(e^t\)):
\(\int t^2 e^t\,dt = e^t(t^2 – 2t + 2)\). Hence:

\[
\int e^{\sqrt[3]{x}}\,dx = 3e^t(t^2 – 2t + 2) + C.
\]

Substituting back \(t = x^{1/3}\):

\[
\int e^{\sqrt[3]{x}}\,dx = 3e^{\sqrt[3]{x}}\!\left(x^{2/3} – 2x^{1/3} + 2\right) + C.
\]