?_R f(x,y)dx dy when converting this into polar form we take x = r cos ?
y=r sin ? as change of variables from
\(\int\int_R f(x,y) \,dx \,dy = \int\int_S f(g(u,v),h(u,v)) \frac{?(x,y)}{?(u,v)} \,du \,dv \) where u=r & v=?
thus \(\frac{?(x,y)}{?(r,?)} =\begin{vmatrix}
\frac{?x}{?r} &\frac{?x}{??}\\
\frac{?y}{?r} &\frac{?y}{??}\\
\end{vmatrix} = \begin{vmatrix}
cos?? &-r sin??\\
sin?? & rcos??\\
\end{vmatrix} = r(cos^2 ? + sin^2 ?) = r\)
substituting we get ?_P f(r cos ?,r sin ?)rdr d?.

Add constant automatically
We know that,
a=\(\int\int[t^2-t] dtdt=\int [\frac{t^3}{3}-\frac{t^2}{2}]dt=\int_0^1\left [\frac{t^3}{3}-\frac{t^2}{2}+10 \right ] dt=\left [\frac{t^4}{12}-\frac{t^3}{6}+10t\right ]\)
=\(\frac{1}{12}-\frac{1}{6}+10=\frac{1}{6} [\frac{1}{2}-1]+10 km=-\frac{1}{12}+10=\frac{119}{12}\)km

Given: \(V = ?_0^2??_0^\frac{?}{3}?_0^1 r cos? \,dr \,d? \,d?.\)
\( V = ?_0^2??_0^{\frac{?}{3}}(\frac{r^2}{2})_0^1 cos?\, d?\, d?\)
\( V = \frac{1}{2} ?_0^{2?}(sin?)_0^\frac{?}{3} d? \,d?\)
\(V = \frac{1}{2}�\frac{?3}{2} ?_0^2? d?\)
\( V = \frac{1}{2}�\frac{?3}{2} �2?\)
\( V = \frac{?3}{2} ? \)

Add constant automatically
Given, \(f(x)=?_0^x (x^2+y^2) \,dxdy = ? (\frac{x^3}{3}+\frac{x^3}{3})\,dxdy = \frac{2x^3}{3} y\).

Add constant automatically
Given,\(\int \frac{1}{16x^2+16x+10} dx=\frac{1}{2}\int \frac{1}{4x^2+4x+5} dx\)
=\(\int \frac{1}{8(x^2+x+5/4+1/4+1/4)} dx\)
=\(\int \frac{1}{8[(x+1/2)^2+1^2]}dx=\frac{1}{8} tan^{-1}?(x+1/2)\)
Hence, \(\frac{1}{8} \int tan^{-1}?(x+\frac{1}{2})dx\)
Now, Putting, x+1/2 = tan(y),
We get, dx = sec²?(y)dy,
=1/8 \(\int ysec^2 (y)dy\)
By integration by parts,
ytan(y)-log?(sec?(y))=\(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln?(1+(x+\frac{1}{2})^2)\)

From figure we can write x = r sin ? cos ?, y = r sin ? sin ?, z = r cos ?
now we know that during a change of variables f(x,y,z) is replaced by \(f(?,?,z)*J\left(\frac{x,y,z}{?,?,z}\right)\) with limits in functions of x,y,z to functions of r,?,? respectively
\(J\left(\frac{x,y,z}{?,?,z}\right) =\begin{vmatrix}
\frac{?x}{?r} &\frac{?x}{??} &\frac{?x}{??}\\
\frac{?y}{?r} &\frac{?y}{??}& \frac{?y}{??}\\
\frac{?z}{?r} &\frac{?z}{??} &\frac{?z}{??}\\
\end{vmatrix} = \begin{vmatrix}
sin ? cos ? &r cos ? cos ? &-r sin ? sin ?\\
sin ? sin ? &r cos ? sin ? &r sin ? cos ?\\
cos ?& -r sin ? &0\\
\end{vmatrix}\)
= sin ? cos ?(r² sin² ? cos??) + r cos ? cos ?(r sin ? cos ? cos ?) – r sin ? sin ?
= (-r sin² ? sin??-r cos² ? sin??)??on solving we get r² sin??
thus ?_R f(x,y,z)dx dy dz = ?_R^* f(r,?,?)r² sin?? dr d? d? where R^* is new region.

W.K.T from change of variables principle
\(\int\int_R f(x,y)\,dx \,dy = \int\int_S f(g(u,v),h(u,v)) \frac{?(x,y)}{?(u,v)} \,du \,dv ?..(1)\)
From above diagram in region R equations are given by
x-y=0, x-y=2, x+y=0, x+y=2 from this we can observe that change of
variables is u=x-y, v=x+y solving we get \(x=\frac{u+v}{2}, y=\frac{v-u}{2}\)
\(\frac{?(x,y)}{?(u,v)} = \begin{vmatrix}
\frac{?x}{?u}&\frac{?x}{?v}\\
\frac{?y}{?u}&\frac{?y}{?v}\\
\end{vmatrix} = \begin{vmatrix}
0.5 &0.5\\
-0.5&0.5\\
\end{vmatrix} = 0.5\)
The region S in (u,v) is square 0 becomes \(\int_0^2 \int_0^2 0.5u^2 \,du \,dv\)??from(1)
\(=\int_0^2 \Big[\frac{u^3}{6}\Big]_0^2 \,dv = \int_0^2 \frac{4}{3} dv = \frac{8}{3}.\)

Equation of circle is given by x² + b² = a²
\(\int_{-a}^a \int_{-a}^a [x^2+y^2-a^2 ]dxdy\)
=\(\int_{-a}^a [\frac{x^3}{3}+y^2 x-a^2 x]_a^{-a}dy=\int_{-a}^a [\frac{a^3}{3}+ay^2-a^2 a+\frac{a^3}{3}+y^2 a-a^2 a]dy\)
\(\int_{-a}^a \frac{2a^3}{3}+2ay^2- 2a^3 dy=[\frac{2a^3 y}{3}+\frac{2ay^3}{3}-2a^3 y] _a^{-a}=\frac{4a^4}{3}+\frac{4a^4}{3}-4a^4=-\frac{4}{3} a^4\)

Add constant automatically
By, f(x)=\(\int uvdx=\sum_{i=0}^n (-1)^i u_i v^{i+1}\),
Let, u = x⁷ and v=Cos(x),
∫x⁷ Cos(x) dx=x⁷ Sin(x)+7x⁶ Cos(x)+42x⁵ Sin(x)+210x⁴ Cos(x)+840x³ Sin(x)+2520x² Cos(x)+5040xSin(x)+5040Cos(x)
Similarly,
∫y⁷ Cos(y) dy=y⁷ Sin(y)+7y⁶ Cos(y)+42y⁵ Sin(y)+210y⁴ Cos(y)+840y³ Sin(y)+2520y² Cos(y)+5040ySin(y)+5040Cos(y)
Now,
∫∫xy⁷ Cos(x)Cos(y) dxdy=∫y⁷ Cos(y) dy∫x⁷ Cos(x) dx=(y⁷ Sin(y)+7y⁶ Cos(y)+42y⁵ Sin(y)+210y⁴ Cos(y)+840y³ Sin(y)+2520y² Cos(y)+5040ySin(y)+5040Cos(y))(x⁷ Sin(x)+7x⁶ Cos(x)+42x⁵ Sin(x)+210x⁴ Cos(x)+840x³ Sin(x)+2520x² Cos(x)+5040xSin(x)+5040Cos(x)).

\(\int_0^2a \int_0^{x^2/4a} x.y dxdy=\int_0^{2a} xdx \int_0^{x^2/4a} ydy=\int_0^{2a}\frac{xx^4}{32a^2}dx\) =\(\frac{1}{32a^2} \int_0^{2a}x^5 dx=\frac{a^4}{3}\)