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Chapter 7 Class 12 Integrals
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Ex 7.7, 1 √(4βˆ’π‘₯2) ∫1β–’γ€–βˆš(4βˆ’π‘₯^2 ).𝑑π‘₯γ€— =∫1β–’γ€–βˆš((𝟐)^πŸβˆ’π’™^𝟐 ).𝒅𝒙〗 =1/2 π‘₯√((2)^2βˆ’π‘₯^2 )+(2)^2/2 𝑠𝑖𝑛^(βˆ’1) π‘₯/2+𝐢 =𝒙/𝟐 √(πŸ’βˆ’π’™^𝟐 )+𝟐 π’”π’Šπ’^(βˆ’πŸ) 𝒙/𝟐+π‘ͺ It is of the form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ). 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 𝑠𝑖𝑛^(βˆ’1) π‘₯/π‘Ž+𝐢〗 Replacing a with 2, we get

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