Misalkan $u=\left(x^{2}+1\right)$

$\frac{du}{dx}=2x$$\rightarrow du=2xdx$

$\begin{aligned}\int2x\cdot cos\,(x^{2}dx) & =\int cosu\, du\\
& =sinu+c\\
& =sin\left(x^{2}+1\right)+c
\end{aligned}
$

Misalkan $u=cosx$

$du=-sinx\, dx$$\rightarrow-du=sinx\, dx$

$\begin{aligned}\int cos^{2}x\, sinx\, dx & =\int u^{2}\cdot-du\\
& =-\frac{1}{3}u^{3}+c\\
& =-\frac{1}{3}cos^{3}x+c
\end{aligned}
$

Misalkan $u=\sqrt{x}$

$du=\frac{1}{2\sqrt{x}}\, dx$$\rightarrow2du=\frac{1}{\sqrt{x}}dx$

$\begin{aligned}\int\frac{cos\sqrt{x}}{\sqrt{x}}\, dx & =\int2\, cosu\, du\\
& =2\, sinu+c\\
& =2\, sin\sqrt{x}+c
\end{aligned}
$

Misalkan $u=cos2x$

$du=-2\, sin2x\, dx$$\rightarrow-\frac{1}{2}du=sin2x\, dx$

$\begin{aligned}\int cos^{4}2x\, sin2x\, dx & =\int-\frac{1}{2}u^{4}du\\
& =-\frac{1}{10}u^{5}+c\\
& =-\frac{1}{10}cos^{5}2x+c
\end{aligned}
$

Misalkan $u=sinx$

$du=cosx\, dx$

$\begin{aligned}\int\frac{cosx}{2sin^{6}x}\, dx & =\int\frac{1}{2u^{6}}\, du\\
& =\frac{1}{2}\int u^{-6}du\\
& =-\frac{1}{10}u^{-5}+c\\
& =-\frac{1}{10sin^{5}x}+c
\end{aligned}
$

Misalkan $u=sin7x$

$du=7cos7x\, dx$

$\frac{1}{2}du=\frac{7}{2}\, cos7x\, dx$$=3,5\, cos7x\, dx$

$\begin{aligned}\int3,5\, cos7x\cdot\,\sqrt{sin^{3}7x}\, dx & =\int\frac{1}{2}u^{\frac{3}{2}}du\\
& =\frac{1}{5}u^{\frac{5}{2}}+c\\
& =\frac{1}{5}\, sin^{\frac{5}{2}}7x+c\\
& =\frac{1}{5}\sqrt{sin^{5}7x}+c
\end{aligned}
$

Misalkan $u=cos2x$

$du=-2\, sin2x\, dx$$\rightarrow-\frac{1}{2}du=sin2x\, dx$

$\begin{aligned}\int\frac{sin2x}{cos^{3}2x}\, dx & =\int-\frac{1}{2}\cdot\frac{1}{u^{3}}\, du\\
& =\frac{1}{4}u^{-2}+c\\
& =\frac{1}{4u^{2}}+c\\
& =\frac{1}{4cos^{2}2x}+c
\end{aligned}
$

$\begin{aligned}\int_{0}^{\frac{\pi}{6}}\frac{sin2x}{cos^{3}2x}\, dx & =\left[\frac{1}{4cos^{2}2x}\right]_{0}^{\frac{\pi}{6}}\\
& =\left(\frac{1}{4cos^{2}2\left(\frac{\pi}{6}\right)}\right)-\left(\frac{1}{4cos^{2}0}\right)\\
& =1-\frac{1}{4}=\frac{3}{4}
\end{aligned}
$

Misalkan : $u=sin\left(2x^{2}+2x\right)$

$du=(4x+2)cos\left(2x^{2}+2x\right)dx$

$\rightarrow5du=\left(20x+10\right)\, cos\left(2x^{2}+2x\right)dx$

$\int\left(20x+10\right)sin^{9}\left(2x^{2}+2x\right)dx$$=\int5u^{9}du$

$=\frac{5}{10}u^{10}+c$

$=\frac{1}{2}u^{10}+c$

$=\frac{1}{2}sin^{10}\left(2x^{2}+2x\right)+c$

Misalkan $u=cos\left(\sqrt{x^{2}+1}\right)$

$du=\frac{2x}{\sqrt{x^{2}+1}}\cdot-sin\left(\sqrt{x^{2}+1}\right)dx$

$-du=\frac{2x}{\sqrt{x^{2}+1}}\cdot sin\left(\sqrt{x^{2}+1}\right)dx$

$\int\frac{2x}{\sqrt{x^{2}+1}}\cdot sin\left(\sqrt{x^{2}+1}\right)\cdot cos^{3}\left(\sqrt{x^{2}+1}\right)\, dx$$=\int-u^{3}du$

$=-\frac{1}{4}u^{4}+c$

$=-\frac{1}{4}cos^{4}\left(\sqrt{x^{2}+1}\right)+c$

Misalkan $u=tan10x$

$du=10sec^{2}10x\, dx$

$\rightarrow5du=50sec^{2}10x\, dx$

$\begin{aligned}\int5sec^{2}\,10x\cdot tan^{4}10x\, dx & =\int5u^{4}du\\
& =u^{5}+c\\
& =tan^{5}10x+c
\end{aligned}
$

Tulis $x=8\,\sin u$, maka $dx=8\cos u\, du$.

Jadi;

$\int\sqrt{64-x^{2}}dx$$=\int\sqrt{64-64\sin^{2}(u)}8\cos(u)du$

$=\int8\sqrt{1-\sin^{2}(u)}8\cos(u)du$

$=\int64\cos^{2}(u)du$

$=32\int(1+\cos(2u))du$

$=32u+16\sin(2u)+c$

Karena $u=\sin^{-1}(\frac{x}{8})$,

maka;

$\int\sqrt{64-x^{2}}dx$$=32\sin^{-1}(\frac{x}{8})$$+16\sin(2\sin^{-1}(\frac{x}{8}))+c$

$=32\sin^{-1}(\frac{x}{8})$$+32\sin(\sin^{-1}(\frac{x}{8}))\,\cos(\sin^{-1}(\frac{x}{8}))+c$

$=32\sin^{-1}(\frac{x}{8})+32(\frac{x}{8})\cos(\sin^{-1}(\frac{x}{8}))+c$

$=32\sin^{-1}(\frac{x}{8})+4x\cos(\sin^{-1}(\frac{x}{8}))+c$

Terakhir, perhatikan bahwa $\cos(\sin^{-1}(a))=\sqrt{1-a^{2}}$,

sehingga;

$\int\sqrt{64-x^{2}}dx$$=32\sin^{-1}(\frac{x}{8})+4x\sqrt{1-\frac{x^{2}}{64}}+c$

$=32\sin^{-1}(\frac{x}{8})+\frac{1}{2}x\sqrt{64-x^{2}}+c$

Tulis $x=2\sin z$, maka $dx=2\cos z\, dz$.

Jadi;

$\begin{aligned}\int\frac{1}{\sqrt{4-x^{2}}}\, dx & =\int\frac{1}{\sqrt{4-4\sin^{2}(z)}}2\cos z\, dz\\
& =\int\frac{1}{2\cos z}2\cos z\, dz\\
& =\int dz\\
& =z+c\\
& =\sin^{-1}(\frac{x}{2})+c
\end{aligned}
$

Tulis $x=3\sin z$, maka $dx=3\cos z\, dz$.

Jadi;

$\begin{aligned}\int\frac{\sqrt{9-x^{2}}}{x^{2}}\, dx & =\int\frac{\sqrt{9-9\sin^{2}z}}{9\sin^{2}z}3\cos z\, dz\\
& =\int\frac{3\cos^{2}z}{9\sin^{2}z}3\cos z\, dz\\
& =\int\frac{\cos^{2}(z)}{\sin^{2}(z)}\, dz\\
& =\int\cot^{2}(z)\, dz\\
& =\int(\csc^{2}(z)-1)\, dz\\
& =\int\csc^{2}(z)dz-\int dz\\
& =-\cot(z)-z+c
\end{aligned}
$

Sekarang, jika $\sin z=\frac{x}{3}$, maka;

$\begin{aligned}\cot^{2}(z) & =\csc^{2}(z)-1\\
& =\frac{1}{\sin^{2}(z)}-1.
\end{aligned}
$

Sehingga;

$\begin{aligned}\cot^{2}(z) & =\frac{1}{(\frac{x}{3})^{2}}-1\\
& =\frac{9}{x^{2}}-1\\
& =\frac{9-x^{2}}{x^{2}}.
\end{aligned}
$

Jadi, $\cot(z)=\frac{\sqrt{9-x^{2}}}{x}$.

Jadi, $\int\frac{\sqrt{9-x^{2}}}{x^{2}}dx=-\frac{\sqrt{9-x^{2}}}{x}-\sin^{-1}(\frac{x}{3})+c.$

Tulis $x=4\sin z$, maka $dx=4\cos z\, dz$.

Jadi

$\int\frac{x^{2}}{\sqrt{16-x^{2}}}\, dx$$=\int\frac{16\sin^{2}z}{\sqrt{16-16\sin^{2}z}}4\cos(z)\, dz$

$=\int\frac{16\sin^{2}z}{\sqrt{16\cos^{2}(z)}}4\cos(z)\, dz$

$=\int16\sin^{2}z\, dz$

$=8(\int1-\cos(2z)\, dz)$

$=8z-4\sin(2z)+c$

$=8\sin^{-1}(\frac{x}{4})-8\sin(z)\cos(z)+c$

$=8\sin^{-1}(\frac{x}{4})-8\sin(z)\sqrt{1-\sin^{2}(z)}+c$

$=8\sin^{-1}(\frac{x}{4})-2x\sqrt{1-\frac{x^{2}}{16}}+c$

$=8\sin^{-1}(\frac{x}{4})-\frac{x}{2}\sqrt{16-x^{2}}+c$

Perhatikan bahwa $=-4+4x-x^{2}+4$$=-(x-2)^{2}+4.$

Tulis $(x-2)=2\sin z$, sehingga $dx=2\cos(z)dz$.

Jadi;

$\int\sqrt{4x-x^{2}}\, dx$$=\int\sqrt{4-(x-2)^{2}}\, dx$

$=\int\sqrt{4-4\sin^{2}(z)}2\cos(z)\, dz$

$=\int4\cos^{2}(z)\, dz$

$=\int2+2\cos(2z)\, dz$

$=2z+\sin(2z)+c$

$=2\sin^{-1}(\frac{x-2}{2})+2\sin(z)\cos(z)+c$

$=2\sin^{-1}(\frac{x-2}{2})+2\sin(z)\sqrt{1-\sin^{2}(z)}+c$

$=2\sin^{-1}(\frac{x-2}{2})+(x-2)\sqrt{1-\frac{(x-2)^{2}}{4}}+c$

$=2\sin^{-1}(\frac{x-2}{2})+\frac{x-2}{2}\sqrt{-(x-2)^{2}+4}+c$

$=2\sin^{-1}(\frac{x-2}{2})+\frac{x-2}{2}\sqrt{4x-x^{2}}+c$