Advertisements
Advertisements
Question
Evaluate `∫_0^(3/2)|x cosπx|dx`
Solution
`int_0^(3/2)|xcospix|dx`
`0<x<1/2`
`0<pix<pi/2rArrcospix>0rArr(xcospix)>0`
`|xcospix|=xcospix`
`1/2<x<3/2`
`pi/2<pix<(3pi)/2rArrcospix<0rArr(xcospix)<0`
`|xcospix|=-xcospix`
`I=int_0^(3/2)|xcospix|dx=int_0^(3/2)xcospix+int_(1/2)^(3/2)-(xcospix)`
`I=int_0^(1/2)xcospix-int_(1/2)^(3/2)xcospix`
`intx(cospix)=x(sinpix)/pi-int(sinpix)/pi`
`=x/pi(sinpix)+(cospix)/pi^2`
`I=[(x/pisinpix)+(cospix)/pi^2]_0^(1/2)-[(x/pisinpix)+(cospix)/pi^2]_(1/2)^(3/2)`
`=[1/pi((1/2)-0)+1/pi^2(0-1)]-[1/pi(3/2(-1)-1/2(1))+1/pi^2(0-0)]`
`=(1/(2pi)-1/pi^2)-((-2)/pi)`
`=(5/(2pi)-1/pi^2)`
`=((5pi-2)/(2pi^2))`
RELATED QUESTIONS
Evaluate : `int1/(3+5cosx)dx`
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate the integral by using substitution.
`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate the integral by using substitution.
`int_0^(pi/2) (sin x)/(1+ cos^2 x) dx`
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
Find: `int_ (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.
