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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.