# how can I plot two variables (inside a definite integral) against each other

I want to plot x against y in this relation: • Your variable h is not defined, and you need to capitalize Exp (not exp). – Roman Jul 1 at 11:09
• 1- it is type mistake, y is true (not h) 2- ok thanks I correct Exp. – R123 Jul 1 at 11:22

You can try ContourPlot.

For example:

ContourPlot[Evaluate@Integrate[Sin[x  y  + x + (y - Pi/2) x  z], {z, 0, 2 Pi}],
{x, -Pi,  Pi}, {y, -Pi, Pi}, PlotRange -> Full, Exclusions -> None] To show a single contour, say the contour where Integrate[Sin[x y + x + (y - Pi/2) x z], {z, 0, 2 Pi}] == Pi/2 you can use

ContourPlot[Evaluate[Integrate[Sin[x  y  + x + (y - Pi/2) x  z], {z, 0, 2 Pi}] == Pi/2],
{x, -Pi, Pi}, {y, -Pi, Pi}, Exclusions -> None,
ContourStyle -> Directive[Red, Thick], ContourShading -> None,
PlotRange -> Full] Alternatively, you can use the option Contours -> {Pi/2}:

ContourPlot[Evaluate@Integrate[Sin[x  y  + x + (y - Pi/2) x  z], {z, 0, 2 Pi}],
{x, -Pi,  Pi}, {y, -Pi, Pi}, Exclusions -> None, ContourStyle -> Thick,
ContourShading -> None, PlotRange -> Full, Contours -> {Pi/2}]


same picture

• Thanks kglr. I will try it. but I have a question. What about the right side of this relation? The left side is equals to 2pi (=2pi). It seems that this is an integral equation. – R123 Jul 1 at 9:47
• @Rojan, if you Integrate[f[x,y,z], {z, 0,2pi}] is a function of (x,y). If you call that function g[x,y], you can use ContourPlot[ g[x,y] == level, {x,...}, {y...}] to get the locus of {x,y} points that satisfy g[x,y] == level`. – kglr Jul 1 at 10:04