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Ex 7.3, 20 - Integrate cos 2x / (cos x + sin x)^2 - NCERT Maths

Ex 7.3, 20 - Chapter 7 Class 12 Integrals - Part 2

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Ex 7.3, 20 Integrate the function cos⁑2π‘₯/(cos⁑〖π‘₯ γ€—+ sin⁑π‘₯ )^2 ∫1β–’cos⁑2π‘₯/(cos⁑π‘₯ + sin⁑π‘₯ )^2 =∫1β–’(cos^2⁑π‘₯ βˆ’ sin^2⁑π‘₯)/(cos⁑π‘₯ + sin⁑π‘₯ )^2 𝑑π‘₯ =∫1β–’(cos⁑π‘₯ βˆ’ sin⁑π‘₯ )(cos⁑π‘₯ + sin⁑π‘₯ )/(cos⁑π‘₯ + sin⁑π‘₯ )^2 𝑑π‘₯ =∫1β–’(cos⁑π‘₯ βˆ’ sin⁑π‘₯)/(cos⁑π‘₯ + sin⁑π‘₯ ) 𝑑π‘₯ Let cos⁑π‘₯+sin⁑π‘₯=𝑑 Differentiating w.r.t. x (π‘π‘œπ‘ β‘2πœƒ=γ€–π‘π‘œπ‘ γ€—^2β‘πœƒβˆ’γ€–π‘ π‘–π‘›γ€—^2β‘πœƒ) βˆ’sin⁑π‘₯+cos⁑π‘₯=𝑑𝑑/𝑑π‘₯ (cos⁑π‘₯βˆ’sin⁑π‘₯ )𝑑π‘₯=𝑑𝑑 𝑑π‘₯=1/((cos⁑π‘₯ βˆ’ sin⁑π‘₯ ) ) 𝑑𝑑 Thus, our equation becomes =∫1β–’((cos⁑π‘₯ βˆ’ sin⁑π‘₯ ))/𝑑 ×𝑑𝑑/(cos⁑π‘₯ βˆ’ sin⁑π‘₯ ) =∫1β–’1/𝑑 𝑑𝑑 =log⁑|𝑑|+𝐢 Putting value of 𝑑=π‘π‘œπ‘ β‘π‘₯+𝑠𝑖𝑛⁑π‘₯ =π’π’π’ˆβ‘|𝒄𝒐𝒔⁑𝒙+π’”π’Šπ’β‘π’™ |+π‘ͺ

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