`int _0^(pi/4) (xsinx)/(cos^3(x))dx`

First the indefinite integral has to be computed. The above integral can be refined as:

`int xtanxsec^2x dx`

Now applying integration by parts:

`int uv'=uv-int u'v `

`u=x, u'=1, v'=sec^2xtanx, v=(sec^2x)/2`

`=x(sec^2x)/2- int 1(sec^2x)/2 dx`

`=(xsec^2x)/2- int (sec^2x)/2 dx`

`=(xsec^2x)/2-(tanx)/2+C`

Computing the boundaries:

`int_0^(pi/4) (xsinx)/(cos^3x) dx=(pi-2)/4-0=(pi-2)/4`

** Therefore,...**

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`int _0^(pi/4) (xsinx)/(cos^3(x))dx`

First the indefinite integral has to be computed. The above integral can be refined as:

`int xtanxsec^2x dx`

Now applying integration by parts:

`int uv'=uv-int u'v `

`u=x, u'=1, v'=sec^2xtanx, v=(sec^2x)/2`

`=x(sec^2x)/2- int 1(sec^2x)/2 dx`

`=(xsec^2x)/2- int (sec^2x)/2 dx`

`=(xsec^2x)/2-(tanx)/2+C`

Computing the boundaries:

`int_0^(pi/4) (xsinx)/(cos^3x) dx=(pi-2)/4-0=(pi-2)/4`

**Therefore, the answer is** `(pi-2)/4` .

**Further Reading**