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प्रश्न
`int cos^7 x "d"x`
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उत्तर
Let I = `int cos^7 x "d"x`
= `int(cos^2x)^3*cosx "d"x`
= `int (1 - sin^2x)^3* cosx "d"x`
Put sin x = t
∴ cos x dx = dt
∴ I = `int (1 - "t"^2)^3 "dt"`
= `int (1 - 3"t"^2 + 3"t"^4 - "t"^6) "dt"`
= `int 1* "dt" - 3 int "t"^2 "dt" + 3 int "t"^4 "dt" - int "t"^6 "dt"`
= `"t" - 3 ("t"^3/3) + 3"t"^5/5) - "t"^7/7 + "c"`
∴ I = `sinx - sin^3x + 3/5 sin^5x - 1/7 sin^7x + "c"`
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