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TARGET PUBLICATION-INTEGRATION-COMPETITIVE THINKING
- intcosalpha/(sinxcos(x-alpha))dx=""+c, " where " 0 le x le alpha le pi...
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- intsec^(5)xtanxdx=?
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- Let I(n)=inttan^(n)xdx , n gt 1. I(4)I(6)=atan^(5)x+bx^(5)+C , wher...
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- For n ge 2, " if " I(n)=intsec^(n)xdx, " then " I(4)-2/3I(2)=
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- int(cos(logex))/xdx
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- intxe^(x^(2)log2)e^(x^(2))dx=""+c
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- Evaluate : int(2"x"dot"tan"^(-1)x)/(1+"x"^4)"\ dx"
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- int (e^(x)+e^(-x))^(2)*(e^(x)-e^(-x))dx is equal to
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- Evaluate: int(1+tanx)/(x+logsecx)dx
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- To find the value of int(1+logx)/(x)dx, the proper substitution is
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- Evaluate : int(sin2x)/(a^2+b^2sin^2x)dx
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- intx^(2)secx^(3)dx=
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- int(e^(x)(1+x))/(cos^(2)(xe^(x)))dx=
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- int(1+x)/(x+e^(-x))dx is equal to
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- int(((x^2+2)a^((x+tan^(- 1)x)))/(x^2+1))dx=
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- The primitive of 1/(4sqrt(x)+x) w.r.t.x is
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- int(xsinx)/(xcosx-sinx-1)dx=""+c
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- The value of int(dx)/(sinxcosx+3cos^(2)x) is
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- int(cosx)/(sqrt(1+sinx))dx is equal to
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- The value of int("cosec x")/(cos^(2)(1+logtan.(x)/(2)))dx is
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