1/cos y = sec y
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Can you please claify if you mean x=y^2/ pi*cos(x) , or x=y^2/cos(pi), since they are very different sums.
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
y=3cos(x) peroid is 2pie
We have:int int (x * sin(y)) dx dyIntegrate x first:int(x)dx = 1/2 * x2 + CNow integrate sin(y):int(sin(y))dy = -cos(y) + CMultiply:-1/2 * x2 * cos(y) + C
y = sec(x)*cot(x)*cos(x)To solve this trigonometric equation, you need to know these identities:sec(x) = 1/(cos(x))cot(x) = 1/(tan(x)) = (cos(x))/(sin(x))Now substitute these identities into the original equation:y = (1/cos(x))*((cos(x))/(sin(x)))*cos(x)Now cancel out the terms that are similar in the numerator and denominator to leave you with:y = (1/(sin(x)))*cos(x)y = (cos(x))/(sin(x))From the aforementioned known identity, the final simplified trigonometric equation becomes:y = cot(x)