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AAKASH SERIES-INDEFINITE INTEGRALS -EXERCISE -II
- int (x.e^(x))/((2 + x)^(3)) dx =
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- int (e^(x)(x^(2)+1))/((1+x)^(2))dx=
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- int ((2-sin 2x)/(1-cos 2x))e^(x)dx=
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- int (3-x^(2))/(1-2x+x^(2))e^(x)dx=e^(x)f(x)+c rArr f(x)=
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- int e^(x).(x^(3) + x + 1)/((1 + x^(2))^(3//2)) dx =
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- int [ sin(log x)+ cos (log x ) ] dx =
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- int [ (1)/(log x) - (1)/((log x)^(2)) ] dx =
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- int [ (log x - 1)/(1 + (log x)^(2)) ]^(2) dx =
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- int e^(x) [ "In " x + (1)/(x^(2)) ] dx =
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- int [ log(log x) + (log x )^(-2) ] dx =
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- int e^(Tan^(-1))((1+x+x^(2))/(1+x^(2)))dx=
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- int e^(sin^(-1)x) { 1 + (x)/(sqrt(1 - x^(2))) } dx =
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- int (x + sin x)/(1 + cosx ) dx =
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- int(x-sinx)/(1+cosx)dx= tan((x)/(2))+plog|sec((x)/(2))|+c rArr p=
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- int (("cosx")/(x) - "sinx.logx") dx =
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- int (1 + x -x^(-1)) e^(x + x^(-1)) dx =
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- int e^(-5x).cos 12 x dx =
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- int 3^(x) cos 5x dx =
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- Evaluate the integerals. int cos (log x) dx on (0,oo).
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- If I(n)= int(sin nx)/(cos x)dx, then I(n)=
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