int(0)^(pi)xf(sinx)dx=...

`int_(0)^(pi)xf(sinx)dx=`

`(pi)/(2)int_(0)^(pi//2)f(sinx)dx`

`piint_(0)^(pi//2)f(cosx)dx`

`piint_(0)^(pi)f(cosx)dx`

`piint_(0)^(pi)f(sinx)dx`

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int(0)^(2)(2x-2)/(2x-x^(2))dx=
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int(0)^(pi)(x.sin2x.sin[(pi//2)cosx])/(2x-pi)dx=
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int(0)^(pi)xf(sinx)dx=
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If int(0)^(pi)xf(sinx)dx=Aint(0)^(pi//2)(sinx)dx, then A is
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If f(x) is intergrable on [0,a] then int(0)^(a)(f(x))/(f(x)+f(a-x))dx=
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int(a)^(b)(f(x))/(f(x)+f(a+b-x))dx=
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If f(a+b-x)=f(x)," then "int(a)^(b)xf(x)dx=
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int(2)^(3)(sqrtx)/(sqrt(5-x)+sqrtx)dx
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The value of the integral, int(3)^(6)(sqrtx)/(sqrt(9-x)+sqrtx)dx is
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int(-a)^(b)(root(3)(x+a))/(root(3)(x+a)+root(3)(b-x))dx=
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If f(x)=(e^(x))/(1+e^(x)),I(1)=int(f(-a))^(f(a))xg{x(1-x)}dxandI(2)=in...
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Let l=int(0)^(1)(sinx)/(sqrtx)dxandI=int(0)^(1)(cosx)/(sqrtx)dx. Then ...
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int(pi//8)^(3pi//8)(1)/(1+cotx)dx=
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int(-2)^(2)(x^(11)cosx+e^(x))dx=
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int(-1)^(1)log((2+x)/(2-x))dx=
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int(-pi//2)^(pi//2)log((2-sintheta)/(2+sintheta))d""theta=
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int(-pi//4)^(pi//4)((x+pi//4)/(2-cos2x))dx=
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int(-1)^(+1)(d)/(dx)(Tan^(-1)1//x)dx=
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The value of int(-pi//2)^(pi//2)(x^(3)+xcosx+tan^(5)x+1)dx=
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int(-3pi//2)^(-pi//2)[(x+pi)^(3)+cos^(2)(x+3pi)]dx=
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