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प्रश्न
Solve: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`. [Hint: Substitute x + y = z]
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उत्तर
Given that: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`
Put x + y = v, on differentiating w.r.t. x, we get,
`1 + ("d"y)/("d"x) = "dv"/"dx"`
∴ `("d"y)/("d"x) = "dv"/"dx" - 1`
∴ `"dv"/"dx" - 1` = cos v + sin v
⇒ `"dv"/"dx"` = cos v + sin v + 1
⇒ `"dv"/(cos"v" + sin"v" + 1)` = dx
Integrating both sides, we have
`int "dv"/(cos"v" + sin"v" + 1) = int 1 . "d"x`
⇒ `int "dv"/(((1 - tan^2 "v"/2)/(1 + tan^2 "v"/2) + (2tan "v"/2)/(1 + tan^2 "v"/2) + 1)) = int 1. "d"x`
⇒ `int ((1 + tan^2 "v"/2))/(1 - tan^2 "v"/2 + 2 tan "v"/2 + 1 + tan^2 "v"/2) "dv" = int 1."d"x`
⇒ `int (sec^2 "v"/2)/(2 + 2 tan "v"/2) "dv" = int 1."d"x`
Put `2 + 2 tan "v"/2` = t
`2 * 1/2 sec^2 "v"/2 "dv"` = dt
⇒ `sec^2 "v"/2 "dv"` = dt
⇒ `int "dt"/"t" = int 1."d"x`
⇒ `log|"t"|` = x + c
⇒ `log|2 + 2 tan "v"/2|` = x + c
⇒ `log|2 + 2tan((x + y)/2)| ` = x + c
⇒ `log2 [1 + tan((x + y)/2)]` = x + c
⇒ `log2 + log[1 + tan ((x + y)/2)]` = x + c
⇒ `log[1 + tan((x + y)/2)]` = x + c – log 2
Hence, the required solution is `log[1 + tan((x + y)/2)]` = x + K ....[c – log 2 = K]
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