HSC Arts कक्षा ११ - Tamil Nadu Board of Secondary Education Question Bank Solutions for Mathematics

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The gradient (slope) of a curve at any point (x, y) is `(x^2 - 4)/x^2`. If the curve passes through the point (2, 7), then the equation of the curve is

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`int ("e"^x (1 + x))/(cos^2(x"e"^x))  "d"x` is

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`int sqrt(tanx)/(sin2x)  "d"x` is

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`int sin^2x "d"x` is

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`int ("e"^(6 log x) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx))  "d"x` is

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`int secx/sqrt(cos2x)  "d"x` is

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`int tan^-1 sqrt((1 - cos 2x)/(1 + cos 2x))  "d"x` is

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`int 2^(3x + 5)  "d"x` is

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`int ("e"^x(x^2 tan^-1x + tan^-1x + 1))/(x^2 + 1)  "d"x` is

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`int (x^2 + cos^2x)/(x^2 + 1)  "cosec"^2 x/("d"x)` is

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`int x^2 cos x  "d"x` is

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`int sqrt((1 - x)/(1 + x))  "d"x` is

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`int ("d"x)/("e"^x - 1)` is

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`int "e"^(- 4x) cos   "d"x` is

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`int (sec^2x)/(tan^2 x - 1)  "d"x`

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`int "e"^(- 7x) sin 5x  "d"x` is

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`int x^2 "e"^(x/2)  "d"x` is

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`int (x + 2)/sqrt(x^2 - 1)  "d"x` is

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`int 1/(x sqrt(log x)^2 - 5)  "d"x` is

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`int sin sqrt(x)  "d"x` is