Exercises — U-substitution — technique, change of limits for definite integrals

Choose $u=x^3+7$. Why? $(x^3+7)$ sits inside the 4th power, and its derivative is $3x^2$ — which is standing right there in front. Differentiate: $du=3x^{2}dx$. Why? This lets the whole clump $3x^{2}dx$ become simply $du$. Replace: $\int u^{4}du$. Every $x$ is gone — including $dx$. Integrate: $\frac{u^5}{5}+C$. Back-substitute (indefinite, so restore $x$): $\boxed{\frac{(x^3+7{)}^5}{5}+C}$.

Choose $u=5x$. Why? The argument of cosine is $5x$; that inner linear function is the thing to rename. Differentiate: $du=5dx\Rightarrow dx=\frac{1}{5}du$. Why? We only have a bare $dx$, so we express it in $u$. Replace: $\int \cos u\cdot \frac{1}{5}du=\frac{1}{5}\int \cos udu$. Integrate: $\frac{1}{5}\sin u+C$. Back-substitute: $\boxed{\frac{1}{5}\sin (5x)+C}$.

Choose $u=x^2+1$. Why? Its derivative $2x$ is exactly the numerator — the classic "$\int \frac{u^{\prime }}{u}$" shape, which produces a logarithm. Differentiate: $du=2xdx$ — the whole numerator-times-$dx$. Replace: $\int \frac{1}{u}du$. Integrate: $\ln |u|+C$. Why the absolute value? $\int \frac{1}{u}du=\ln |u|$ works for negative $u$ too; here $x^2+1>0$ always, so $|u|=u$. Back-substitute: $\boxed{\ln (x^2+1)+C}$.

Choose $u=x^2-4$, so $du=2xdx$. Why? Inner function is $x^2-4$; its derivative is $2x$, and we have $x$ — proportional, not equal. Fix the constant: we have $xdx$ but need $2xdx$, so $xdx=\frac{1}{2}du$. This is legal because $\frac{1}{2}$ is a constant and $\int$ is linear over constants. Replace: $\int u^3\cdot \frac{1}{2}du=\frac{1}{2}\int u^{3}du$. Integrate: $\frac{1}{2}\cdot \frac{u^4}{4}=\frac{u^4}{8}$. Back-substitute: $\boxed{\frac{(x^2-4{)}^4}{8}+C}$.

Choose $u=x^2$, so $du=2xdx\Rightarrow xdx=\frac{1}{2}du$. Change limits (Why? the variable is $u$ now, so the old numbers $0,1$ no longer describe the range):

• $x=0\Rightarrow u=0^2=0$
• $x=1\Rightarrow u=1^2=1$

Read the figure below. The left panel (blue) is the original region: the area trapped under $y=x{e}^{x^2}$ as $x$ walks from $0$ to $1$. The curve starts flat near $x=0$ (because the factor $x$ kills the height there) and shoots up near $x=1$. The right panel (orange) is the same total area after the substitution: we now plot $y=\frac{1}{2}{e}^u$ as $u$ walks from $0$ to $1$. The substitution has stretched and reshaped the horizontal axis — points that were bunched near $x=0$ get spread out in $u$ — yet the shaded area is identical. That is what "u-sub conserves area" means: the $\frac{1}{2}$ and the $du=2xdx$ bookkeeping exactly compensate for the axis being warped. The two endpoints $0$ and $1$ happen to stay put here only because $g(0)=0$ and $g(1)=1$; the interior of the region still transforms.

Replace: ${\int }_0^1{e}^u\cdot \frac{1}{2}du=\frac{1}{2}{\int }_0^1{e}^{u}du$. Integrate & evaluate: $\frac{1}{2}[e^u{]}_0^1=\frac{1}{2}(e^1-e^0)=\frac{1}{2}(e-1)$. $\boxed{\frac{e-1}{2}\approx 0.859}$.

Choose $u=\ln x$, so $du=\frac{1}{x}dx$. Why? The factor $\frac{1}{x}dx$ is exactly $du$, and $\ln x$ appears alone — a perfect "$u$ and $du$" pair. Change limits:

• $x=1\Rightarrow u=\ln 1=0$
• $x=2\Rightarrow u=\ln 2\approx 0.693$

Replace: ${\int }_0^{\ln 2}udu$. Integrate & evaluate: $[\frac{u^2}{2}{]}_0^{\ln 2}=\frac{(\ln 2{)}^2}{2}$. $\boxed{\frac{(\ln 2{)}^2}{2}\approx 0.240}$.

Choose $u=x+1$, so $du=dx$. Why? The awkward part is the root; its inside is $x+1$. Problem: the extra factor $x$ is a variable, not a constant — we cannot pull it out. Instead, since $u=x+1$ means $x=u-1$, we replace that too. Replace: $\int (u-1)\sqrt{u}du=\int (u^{3/2}-u^{1/2})du$. Integrate: $\frac{u^{5/2}}{5/2}-\frac{u^{3/2}}{3/2}=\frac{2}{5}{u}^{5/2}-\frac{2}{3}{u}^{3/2}$. Back-substitute: $\boxed{\frac{2}{5}(x+1{)}^{5/2}-\frac{2}{3}(x+1{)}^{3/2}+C}$.

Rewrite first: $\tan x=\frac{\sin x}{\cos x}$. Why? Now the denominator $\cos x$ has derivative $-\sin x$, which is (up to sign) the numerator — the "$\int \frac{u^{\prime }}{u}$" shape appears. Choose $u=\cos x$, so $du=-\sin xdx\Rightarrow \sin xdx=-du$. Replace: $\int \frac{1}{u}\cdot (-du)=-\int \frac{1}{u}du$. Integrate: $-\ln |u|+C$. Back-substitute: $\boxed{-\ln |\cos x|+C}$ (equivalently $\ln |\sec x|+C$). Why the absolute value stays now? $\cos x$ is negative on some intervals, so $|u|$ is genuinely needed.

Choose $u=\sin x$, so $du=\cos xdx$. Why? $\cos xdx$ is exactly $du$, leaving $\sin x=u$ alone. Change limits:

• $x=0\Rightarrow u=\sin 0=0$
• $x=\pi /2\Rightarrow u=\sin (\pi /2)=1$

Replace: ${\int }_0^{1}udu$. Integrate & evaluate: $[\frac{u^2}{2}{]}_0^1=\frac{1}{2}$. $\boxed{\frac{1}{2}}$. Cross-check the other choice: $u=\cos x$ gives $du=-\sin xdx$, limits $1\to 0$, integral $-{\int }_1^{0}udu={\int }_0^{1}udu=\frac{1}{2}$. Same answer — a good sanity habit.

Substitute to remove the root: let $u=\sqrt{x}$, so $x=u^2$ and $dx=2udu$. Why? The root inside $e^{\sqrt{x}}$ blocks any simple antiderivative; renaming it clears the exponent. Replace: $\int e^u\cdot 2udu=2\int u{e}^{u}du$. Now Integration by Parts. What is that rule? It reverses the product rule: from $\frac{d}{dx}(pq)=p{q}^{\prime }+q{p}^{\prime }$, integrating both sides gives $\int pdq=pq-\int qdp.$ Why parts here and not more u-sub? The integrand $u{e}^u$ is a product of two unrelated pieces (a polynomial $u$ times an exponential $e^u$); neither is the derivative of the other, so no u-sub can collapse it. Parts is the tool for products. Why choose $p=u$ and $dq=e^{u}du$ (and not the reverse)? We want the surviving integral $\int qdp$ to be simpler. Picking $p=u$ makes $dp=du$ — differentiating $u$ shrinks it to a constant, killing the polynomial. Picking $dq=e^{u}du$ is safe because $e^u$ integrates back to $e^u$ (it never grows worse). So $p=u,dq=e^{u}du\Rightarrow dp=du,q=e^u$, giving $\int u{e}^{u}du=u{e}^u-\int e^{u}du=u{e}^u-e^u.$ Assemble: $2(u{e}^u-e^u)=2e^u(u-1)$. Back-substitute $u=\sqrt{x}$: $\boxed{2e^{\sqrt{x}}(\sqrt{x}-1)+C}$.

Notice: the derivative of $1-x^2$ is $-2x$, and we have an $x$ up top — plain u-sub handles this without any trig. Choose $u=1-x^2$, so $du=-2xdx\Rightarrow xdx=-\frac{1}{2}du$. Replace: $\int \frac{1}{\sqrt{u}}\cdot (-\frac{1}{2})du=-\frac{1}{2}\int u^{-1/2}du$. Integrate: $-\frac{1}{2}\cdot \frac{u^{1/2}}{1/2}=-u^{1/2}$. Back-substitute: $\boxed{-\sqrt{1-x^2}+C}$. Why not trig substitution? Trig sub is for $\int \frac{1}{\sqrt{1-x^2}}$ (no $x$ on top). The stray $x$ here makes it a pure u-sub — always try u-sub first.

Choose $u=1+\tan x$, so $du={\sec }^{2}xdx$. Why? The derivative of $\tan x$ is ${\sec }^{2}x$, sitting right in the numerator — the "$\int \frac{u^{\prime }}{u}$" logarithm pattern. Change limits:

• $x=0\Rightarrow u=1+\tan 0=1$
• $x=\pi /4\Rightarrow u=1+\tan (\pi /4)=1+1=2$

Replace: ${\int }_1^2\frac{1}{u}du$. Integrate & evaluate: $[\ln |u|{]}_1^2=\ln 2-\ln 1=\ln 2$. $\boxed{\ln 2\approx 0.693}$.

Peel one layer: let $u=\ln (\ln x)$. Why the outermost? Its derivative, by the Chain Rule, is $\frac{1}{\ln x}\cdot \frac{1}{x}=\frac{1}{x\ln x}$ — precisely the rest of the integrand. Differentiate: $du=\frac{1}{x\ln x}dx$. Replace: $\int \frac{1}{u}du$. Integrate: $\ln |u|+C$. Back-substitute: $\boxed{\ln |\ln (\ln x)|+C}$.

Domain check first. The root $\sqrt{x^3+1}$ is real only when $x^3+1\ge 0$, i.e. $x\ge -1$. Our interval is $[0,3]$, which sits comfortably inside $x\ge -1$, so the integrand is real and continuous the whole way — no gap under the radical to worry about. Choose $u=x^3+1$, so $du=3x^{2}dx\Rightarrow x^{2}dx=\frac{1}{3}du$. Change limits:

• $x=0\Rightarrow u=0^3+1=1$
• $x=3\Rightarrow u=3^3+1=28$

(Since $x\ge 0$ on our interval, $u=x^3+1\ge 1>0$, so $\sqrt{u}$ is always defined in the new variable too.) Replace: ${\int }_1^{28}\sqrt{u}\cdot \frac{1}{3}du=\frac{1}{3}{\int }_1^{28}{u}^{1/2}du$. Integrate: $\frac{1}{3}\cdot \frac{u^{3/2}}{3/2}=\frac{2}{9}{u}^{3/2}$. Evaluate: $\frac{2}{9}(28^{3/2}-1^{3/2})=\frac{2}{9}(28\sqrt{28}-1)$. Since $28\sqrt{28}=28\cdot 2\sqrt{7}=56\sqrt{7}$, this is $\frac{2}{9}(56\sqrt{7}-1)$. $\boxed{\frac{2(56\sqrt{7}-1)}{9}\approx 32.68}$.

Choose $u=x^4$, so $du=4x^{3}dx\Rightarrow x^{3}dx=\frac{1}{4}du$. Change limits — watch the squaring:

• $x=-1\Rightarrow u=(-1{)}^4=1$
• $x=1\Rightarrow u=1^4=1$

Both endpoints map to $u=1$! Replace: $\frac{1}{4}{\int }_1^1{e}^{u}du=0$. $\boxed{0}$. Why does this make sense? $x^3{e}^{x^4}$ is an odd function (replace $x\to -x$: $(-x{)}^3{e}^{(-x{)}^4}=-x^3{e}^{x^4}$), and the integral of an odd function over a symmetric interval $[-1,1]$ is zero. The substitution collapsing both limits to the same value is the algebra showing that symmetry.

Rewrite the top: $e^{2x}=e^x\cdot e^x=(e^x{)}^2$. Why this rewrite? We plan to set $u=e^x+1$, whose differential is $du=e^{x}dx$. Splitting $e^{2x}$ into two factors of $e^x$ lets one $e^x$ pair up with $dx$ to become $du$, while the other $e^x$ becomes $u-1$ — turning everything into the new variable cleanly. Choose $u=e^x+1$, so $e^x=u-1$ and $du=e^{x}dx$. Replace: the integrand $\frac{e^x\cdot e^x}{e^x+1}dx=\frac{(u-1)}{u}du$ (one $e^x$ became $du$, the other became $u-1$). $\int \frac{u-1}{u}du=\int (1-\frac{1}{u})du=u-\ln |u|.$ Back-substitute $u=e^x+1$: $\boxed{(e^x+1)-\ln (e^x+1)+C}$. Since the constant $1$ merges into $C$, this equals $e^x-\ln (e^x+1)+C$.

• Is my $u$ the inner function whose derivative appears? ✔
• Did I convert every $x$-piece, including $dx\to du$? ✔
• Constant mismatch → fix with a number; variable mismatch → solve $x=g^{-1}(u)$. ✔
• Definite? Change limits or back-substitute — never both. ✔
• Checked the domain under any radical/log, and computed new limits by plugging in? ✔
• Did I sanity-check signs, absolute values in logs, and symmetry? ✔