Exercises — Integration by parts — derivation from product rule, LIATE mnemonic

Types: $x$ is Algebraic, $\sin x$ is Trigonometric. LIATE says A comes before T, so $u=x,dv=\sin xdx.$ Differentiate $u$ and integrate $dv$: $du=dx,v=\int \sin xdx=-\cos x.$ Why this choice: differentiating $x$ turns it into the constant $1$ — the polynomial "disappears," making the new integral simpler. Had we chosen $u=\sin x$, then $du=\cos xdx$ never simplifies and $dv=xdx$ gives $v=\frac{x^2}{2}$, which makes the leftover integral worse.

Types: $\ln x$ is Logarithmic, $x^2$ is Algebraic. L before A, so $u=\ln x,dv=x^{2}dx.$ $du=\frac{1}{x}dx,v=\int x^{2}dx=\frac{x^3}{3}.$ Why: $\ln x$ has no tidy elementary antiderivative to reach for directly, but its derivative $1/x$ is clean. So we differentiate it, never integrate it.

Write the "hidden product" $\arctan x=\arctan x\cdot 1$. Now there are two factors. Types: $\arctan x$ is Inverse-trig, the $1$ is Algebraic. I before A, so $u=\arctan x,dv=1dx.$ $du=\frac{1}{1+x^2}dx,v=\int 1dx=x.$ Why: inverse-trig functions are unpleasant to integrate but have rational derivatives — perfect $u$ material.

$u=x$ (A), $dv=\cos xdx$ (T), so $du=dx$, $v=\sin x$. $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C.$ Verify by differentiating: $\frac{d}{dx}(x\sin x+\cos x)=\sin x+x\cos x-\sin x=x\cos x.$ ✓

$u=x$ (A), $dv=e^{2x}dx$ (E). Then $du=dx$ and $v=\int e^{2x}dx=\frac{1}{2}{e}^{2x}.$ Why the $\frac{1}{2}$: the chain rule means $\frac{d}{dx}{e}^{2x}=2e^{2x}$, so to undo it we divide by $2$. $\int x{e}^{2x}dx=x\cdot \frac{1}{2}{e}^{2x}-\int \frac{1}{2}{e}^{2x}dx=\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C=\frac{1}{4}{e}^{2x}(2x-1)+C.$ Verify: $\frac{d}{dx}[\frac{1}{4}{e}^{2x}(2x-1)]=\frac{1}{4}(2e^{2x}(2x-1)+2e^{2x})=\frac{1}{4}{e}^{2x}(4x)=x{e}^{2x}.$ ✓

Hidden product: $\ln (3x)=\ln (3x)\cdot 1$. Take $u=\ln (3x)$, $dv=dx$. $du=\frac{1}{3x}\cdot 3dx=\frac{1}{x}dx,v=x.$ Why $du=\frac{1}{x}dx$: by the chain rule $\frac{d}{dx}\ln (3x)=\frac{1}{3x}\cdot 3=\frac{1}{x}$. (Also $\ln (3x)=\ln 3+\ln x$, whose derivative is plainly $\frac{1}{x}$.) $\int \ln (3x)dx=x\ln (3x)-\int x\cdot \frac{1}{x}dx=x\ln (3x)-\int 1dx=x\ln (3x)-x+C.$ Verify: $\frac{d}{dx}[x\ln (3x)-x]=\ln (3x)+x\cdot \frac{1}{x}-1=\ln (3x).$ ✓

$u=x^2$ (A), $dv=e^{x}dx$, so $du=2xdx$, $v=e^x$: $\int x^2{e}^{x}dx=x^2{e}^x-\int 2x{e}^{x}dx.$ The leftover $\int 2x{e}^{x}dx$ still has a product, so apply parts again with $u=2x$, $dv=e^{x}dx$ ($du=2dx$, $v=e^x$): $\int 2x{e}^{x}dx=2x{e}^x-\int 2e^{x}dx=2x{e}^x-2e^x.$ Combine (mind the leading minus): $\int x^2{e}^{x}dx=x^2{e}^x-(2x{e}^x-2e^x)+C=e^x(x^2-2x+2)+C.$ Why it terminates: each pass drops the polynomial degree by one; after two passes the $x$ is gone. Verify: $\frac{d}{dx}[e^x(x^2-2x+2)]=e^x(x^2-2x+2)+e^x(2x-2)=e^x{x}^2.$ ✓

Let $I=\int e^x\cos xdx$. Take $u=\cos x$, $dv=e^{x}dx$ ($du=-\sin xdx$, $v=e^x$): $I=e^x\cos x-\int e^x(-\sin x)dx=e^x\cos x+\int e^x\sin xdx.$ Apply parts to the new integral with $u=\sin x$, $dv=e^{x}dx$ ($du=\cos xdx$, $v=e^x$): $\int e^x\sin xdx=e^x\sin x-\int e^x\cos xdx=e^x\sin x-I.$ Substitute back: $I=e^x\cos x+e^x\sin x-I\Rightarrow 2I=e^x(\cos x+\sin x).$ $\boxed{I=\frac{1}{2}{e}^x(\cos x+\sin x)+C}$ Why it works: the original integral $I$ reappears, so we solve for it algebraically rather than integrating forever. (Keep the $u,dv$ choice consistent — both times the exponential is $dv$ — or the loop cancels to $0=0$.) Verify: $\frac{d}{dx}[\frac{1}{2}{e}^x(\cos x+\sin x)]=\frac{1}{2}(e^x(\cos x+\sin x)+e^x(-\sin x+\cos x))=\frac{1}{2}{e}^x(2\cos x)=e^x\cos x.$ ✓

$u=(\ln x{)}^2$ (L), $dv=xdx$. Then $v=\frac{x^2}{2}$ and, by the chain rule, $du=2\ln x\cdot \frac{1}{x}dx=\frac{2\ln x}{x}dx.$ $\int x(\ln x{)}^{2}dx=\frac{x^2}{2}(\ln x{)}^2-\int \frac{x^2}{2}\cdot \frac{2\ln x}{x}dx=\frac{x^2}{2}(\ln x{)}^2-\int x\ln xdx.$ Now do parts on $\int x\ln xdx$: $u=\ln x$, $dv=xdx$, $v=\frac{x^2}{2}$, $du=\frac{1}{x}dx$: $\int x\ln xdx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot \frac{1}{x}dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}.$ Combine: $\int x(\ln x{)}^{2}dx=\frac{x^2}{2}(\ln x{)}^2-\frac{x^2}{2}\ln x+\frac{x^2}{4}+C.$ Verify: differentiating gives $x(\ln x{)}^2$ (checked in VERIFY).

First find the antiderivative (parent Example 1): $\int x{e}^{x}dx=e^x(x-1)$. By the Fundamental Theorem of Calculus, evaluate between the limits: ${\int }_0^{1}x{e}^{x}dx=[e^x(x-1){]}_0^1=e^1(1-1)-e^0(0-1)=0-(1)(-1)=1.$ Answer: $\boxed{1}$. For definite integrals you may also carry the limits through each step directly: $[uv{]}_0^1-{\int }_0^{1}vdu$.

Substitute first to remove the awkward $\sqrt{x}$. Let $t=\sqrt{x}$, so $x=t^2$ and $dx=2tdt$. $\int \cos (\sqrt{x})dx=\int \cos t\cdot 2tdt=2\int t\cos tdt.$ Now this is Exercise 2.1's shape. With $u=t$, $dv=\cos tdt$ ($du=dt$, $v=\sin t$): $2\int t\cos tdt=2(t\sin t-\int \sin tdt)=2(t\sin t+\cos t).$ Back-substitute $t=\sqrt{x}$: $\int \cos (\sqrt{x})dx=2\sqrt{x}\sin (\sqrt{x})+2\cos (\sqrt{x})+C.$ Why substitution first: parts alone cannot handle $\cos (\sqrt{x})$ — there is no clean $u/dv$ split. The substitution turns it into a plain polynomial-times-trig product, which parts eats easily. This is the "try Integration by substitution to set up parts" pattern. Verify: differentiating returns $\cos (\sqrt{x})$ (checked in VERIFY).

$u=\arctan x$ (I), $dv=xdx$. Then $v=\frac{x^2}{2}$, $du=\frac{1}{1+x^2}dx$: $\int x\arctan xdx=\frac{x^2}{2}\arctan x-\int \frac{x^2}{2}\cdot \frac{1}{1+x^2}dx=\frac{x^2}{2}\arctan x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx.$ The leftover integral is a rational one. Split by writing $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$: $\frac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)dx=\frac{1}{2}(x-\arctan x).$ Therefore $\int x\arctan xdx=\frac{x^2}{2}\arctan x-\frac{x}{2}+\frac{1}{2}\arctan x+C.$ Verify: derivative equals $x\arctan x$ (checked in VERIFY).

Take $u=x^n$ (A), $dv=e^{x}dx$, so $du=n{x}^{n-1}dx$, $v=e^x$: $I_n=\int x^n{e}^{x}dx=x^n{e}^x-\int e^x\cdot n{x}^{n-1}dx=x^n{e}^x-n\int x^{n-1}{e}^{x}dx=x^n{e}^x-n{I}_{n-1}.$ This is the reduction formula: each application lowers the power by one. Now unroll from $n=3$, using $I_0=\int e^{x}dx=e^x$: $I_1=x{e}^x-I_0=x{e}^x-e^x=e^x(x-1),$ $I_2=x^2{e}^x-2I_1=x^2{e}^x-2e^x(x-1)=e^x(x^2-2x+2),$ $I_3=x^3{e}^x-3I_2=x^3{e}^x-3e^x(x^2-2x+2)=e^x(x^3-3x^2+6x-6).$ $\boxed{\int x^3{e}^{x}dx=e^x(x^3-3x^2+6x-6)+C.}$ Verify: differentiating gives $x^3{e}^x$ (checked in VERIFY).

Antiderivative from Exercise 2.1: $\int x\sin xdx=-x\cos x+\sin x$. (Same as $x\sin x+\cos x$? No — that was $x\cos x$. For $x\sin x$: $u=x$, $dv=\sin xdx$, $v=-\cos x$, giving $-x\cos x+\int \cos xdx=-x\cos x+\sin x$.) Evaluate on $[0,\pi ]$: ${\int }_0^{\pi }x\sin xdx=[-x\cos x+\sin x{]}_0^{\pi }.$ At $x=\pi$: $-\pi \cos \pi +\sin \pi =-\pi (-1)+0=\pi$. At $x=0$: $-0\cdot \cos 0+\sin 0=0$. ${\int }_0^{\pi }x\sin xdx=\pi -0=\boxed{\pi }.✓$

Tabular integration is repeated parts organised in a table: Differentiate $x^3$ down to $0$, Integrate $e^x$ (which stays $e^x$), and read off products along diagonals with alternating signs $+,-,+,-$:

Multiply each D-row by the I-row one step down, apply the sign: $+x^3{e}^x-3x^2{e}^x+6x{e}^x-6e^x+C=e^x(x^3-3x^2+6x-6)+C.$ Why it matches 5.1: tabular integration is the reduction formula applied repeatedly — the alternating signs are exactly the $-n{I}_{n-1}$ minus signs stacking up. It works whenever one factor differentiates to $0$ after finitely many steps (polynomials) and the other integrates cleanly every time.