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There are some standard substitutions, where we put value of x and solve it

Inverse Trigonometry Substitution - Part 2

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Transcript

Sometimes while integrating, We get questions like ∫1β–’βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯ Now, how do we solve it? There are some standard substitutions, where we put value of x and solve it √(π‘Ž^2βˆ’π‘₯^2 ) π‘₯=π‘Ž π‘ π‘–π‘›β‘πœƒ 1βˆ’γ€–π‘ π‘–π‘›γ€—^2β‘πœƒ=γ€–π‘π‘œπ‘ γ€—^2β‘πœƒ √(π‘Ž^2+π‘₯^2 ) π‘₯=π‘Ž π‘‘π‘Žπ‘›β‘πœƒ 1+γ€–π‘‘π‘Žπ‘›γ€—^2β‘πœƒ=〖𝑠𝑒𝑐〗^2β‘πœƒ √(π‘₯^2βˆ’π‘Ž^2 ) π‘₯=π‘Ž π‘ π‘’π‘β‘πœƒ 〖𝑠𝑒𝑐〗^2β‘πœƒβˆ’1=γ€–π‘‘π‘Žπ‘›γ€—^2β‘πœƒ

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