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Evaluate $\displaystyle{\int \cos dx}$
$\displaystyle{\sin x + C}$
Evaluate $\displaystyle{\int \left( x^2 + 2\sqrt{x} - \frac{3}{x^2} - \frac{1}{x} \right) dx}$
Simply integrate term by term to find
$$\displaystyle{\int \left( x^2 + 2\sqrt{x} - \frac{3}{x^2} - \frac{1}{x} \right) dx = \frac{x^3}{3} + \frac{4}{3} x\sqrt{x} + \frac{3}{x} - \ln|x| + C}$$Evaluate $\displaystyle{\int \cos(3x) \, dx}$
Use the substitution $u = 3x$, so that $du = 3 \, dx$. Thus, $$\int \cos(3x) \, dx = \frac{1}{3} \int (\cos u) \, du = \frac{1}{3} \sin u + C = \frac{\sin(3x)}{3} + C$$
Find the position of a particle if $a(t) = 3t^2$ m/s2, $v(1) = 2$ m/s, and $s(0) = 4$ m.
Note the velocity $v(t)$ can be found with $\int a(t) \, dt$, using $v(1) = 2$ m/s to find the particular constant $C$ needed.
Integrating, we then have $v(t) = t^3 + C$. Using $v(1) = 2$, we have $1^3 + C = 2$, so $C = 1$ and $v(t) = t^3 + 1$.
We do the same to $v(t)$ to find $s(t)$. Integrate $v(t)$ and use $s(0) = 4$ m to find the (new) particular constant $C$ needed.
Integrating, we have $s(t) = \frac{1}{4}t^4 + t + C$. Combining this with $s(0) = 4$ tells us immediately that $C = 4$.
Thus, $\displaystyle{s(t) = \frac{t^4}{4} + t + 4}$
Evaluate $\displaystyle{\int \cos x \sin^2 x \, dx}$
Use $u = \sin x$, which tells us $du = \cos x \, dx$. Thus,
$$\int \cos x \sin^2 x \, dx = \int u^2 \, du = \frac{1}{3} u^3 + C = \frac{1}{3}\sin^3 x + C$$Evaluate $\displaystyle{\int x \sqrt{x^2 + 1} \, dx}$
Use $u = x^2 + 1$, which tells us $du = 2x \, dx$. Thus, $$\int x \sqrt{x^2 + 1} \, dx = \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{(x^2+1)^{3/2}}{3}+C$$
Evaluate $\displaystyle{\int \sin(5x) \, dx}$
Use $u = 5x$, so $du = 5 \, dx$. Thus, $$\int \sin(5x) \, dx = \frac{1}{5} \int \sin u \, du = \frac{-\cos u}{5} + C = -\frac{\cos(5x)}{5}+C$$
Evaluate $\displaystyle{\int \sec^2 (x+5) \, dx}$
Use $u = x+5$, so $du = dx$. Thus, $$\int \sec^2 (x+5) \, dx = \int \sec^2 u \, du = \tan u + C = \tan(x+5) + C$$
Evaluate $\displaystyle{\int 3t^2\sqrt{t^3-2} \, dt}$
Use $u = t^3 - 2$, so $du = 3t^2 \, dt$. Thus, $$\int 3t^2\sqrt{t^3-2} \, dt = \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C = \frac{2(t^3-2)^{3/2}}{3}+C$$
Evaluate $\displaystyle{\int 3e^{3s} \, ds}$
Use $u = 3s$, so $du = 3ds$. Thus, $$\int 3e^{3s} \, ds = \int e^u \, du = e^u + C = e^{3s} + C$$
Evaluate $\displaystyle{\int \frac{s \sin(s^2)}{\sqrt{1+\cos(s^2)}} ds}$.
Use the substitution $\displaystyle{u = 1 + \cos(s^2)}$. This tells us that $du = -\sin(s^2) \cdot 2s$. Thus,
$$\int \frac{s \sin(s^2)}{\sqrt{1+\cos(s^2)}} ds = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} \cdot 2 u^{1/2} + C = -\sqrt{1 + \cos(s^2)} + C$$Evaluate $\displaystyle{\int \frac{t^3 + 2t}{3t^2} \, dt}$
Split the integral up as a sum of two terms, then use the power rule: $$\int (\textstyle{\frac{1}{3}t + \frac{2}{3}t^{-1}}) \, dt = \frac{1}{6}t^2 + \frac{2}{3} \ln |t| + C$$
Evaluate $\displaystyle{\int \csc^2(x) \cot(x) \, dx}$
Use a substitution of $u=\cot x$, which tells us that $du = -\csc^2(x) \, dx$. Thus,
$$\int \csc^2(x) \cot(x) \, dx = - \int u \, du = -\frac{1}{2} u^2 + C = -\frac{1}{2} \cot^2 x + C$$Evaluate $\displaystyle{\int \frac{e^{-y} + e^{3y}}{3e^y} \, dy}$
First, let's try a substitution of $u = e^y$ which implies $du = e^y \, dy$, or equivalently $dy = \frac{du}{u}$.
$$\int \frac{e^{-y} + e^{3y}}{3e^y} \, dy = \int \frac{u^{-1} + u^3}{3u} \cdot \frac{1}{u} du = \int \frac{u^{-1} + u^3}{3u^2} du$$Then, we split the fraction so we can use the power rule for integrals
$$\int \left( \frac{1}{3} u^{-3} + \frac{1}{3} u \right) \, du = \frac{-u^{-2}}{6} + \frac{u^2}{6} + C$$Finally, replacing each $u$ with $e^y$ to express our answer in terms of the original variable $y$, we have our answer:
$$\frac{1}{6}(e^{2y} - e^{-2y}) + C$$Evaluate $\displaystyle{\int \cos \theta \sin \theta\,d\theta}$
$\displaystyle{\frac{\sin^2 \theta}{2} + C}$
Evaluate $\displaystyle{\int \tan^2 u \,du}$
$\displaystyle{\int \tan^2 u \,du = \int (\sec^2 u - 1)\,du = \tan u - u + C}$
Evaluate $\displaystyle{\int \cos^3 2x\,dx}$
$\displaystyle{\int \cos^3 2x\,dx = \int (1-\sin^2 2x)\cos 2x\, dx}$, then let $u=\sin 2x$, which means $du = 2\cos 2x\, dx$.
Then, we arrive at $\displaystyle{\frac{1}{2} \int (1-u^2)\, du = \frac{1}{2} \left( u - \frac{1}{3} u^3 \right) + C = \frac{1}{2} \left(\sin 2x - \frac{1}{3} \sin^3 2x \right) + C}$
Evaluate $\displaystyle{\int \alpha \cot^2 \alpha^2\,d\alpha}$
$\displaystyle{-\frac{1}{2}\cot^2 \alpha^2 - \alpha^2 + C}$
Evaluate $\displaystyle{\int \frac{\cos t \ln \sin t}{\sin t}\,dt}$
$\displaystyle{\frac{1}{2}\left(\ln \sin t\right)^2 + C}$
Evaluate $\displaystyle{\int (x-2)\sqrt{3-4x+x^2}\,dx}$
$\displaystyle{(3-4x+x^2)^{3/2} + C}$
Evaluate $\displaystyle{\int \sin^3 s\,ds}$
$\displaystyle{-\cos s + \frac{1}{2} \cos^3 s + C}$
Evaluate $\displaystyle{\int \sin^3 s \cos s\,ds}$
$\displaystyle{\frac{1}{4} \sin^4 s + C}$
Evaluate $\displaystyle{\int \frac{ds}{\sin^2 s}}$
$\displaystyle{-\cot s + C}$
Evaluate $\displaystyle{\int \sec^3 t \tan t\,dt}$
$\displaystyle{\frac{1}{3}\sec^3 t + C}$
Evaluate $\displaystyle{\int \sec^2 t \tan^2 t\,dt}$
$\displaystyle{ \frac{1}{3}\tan^3 t + C}$
Evaluate $\displaystyle{\int \frac{\ln(-u)}{u}\,du}$
$\displaystyle{\frac{1}{2}\ln^2 (-u) + C}$
Evaluate $\displaystyle{\int \frac{e^x\,dx}{e^x + 4}}$
$\displaystyle{\ln(e^x+4) + C}$