The value of-50-40-1-d- where -30 -a cons-an t - is

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Pythagoras Theorem

the value of∫...

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the value of∫_{50}^{40}(1/θ-θ^°)\operatorname dθ where θ^°=30 (a cons†an t ) is

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\frac{sin θ}{cot θ + csc θ} = 2 + \frac{sin θ}{cot d θ - csc θ}

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\int_{0}^{\frac{π}{4}} s e c^{7} θ s i n^{3} θ d θ =

c o s^{2} θ + 2 s i n^{2} θ + 3 c o s^{2} θ + 4 s i n^{2} θ + . . . . . . (200) t e r m s = 10025

θ

s i n θ - c o s θ

θ

(θ + 36^{\circ}) = cos θ

θ + 36^{\circ}

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