# Determine definite integral integrate from 0 to (1/7) of xarcsin(7x)dx

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### 1 Answer

You should use integration by parts, hence, you need to remember the following formula such that:

`int fdg = fg - int gdf`

You need to select `f(x) = arcsin 7x` and `dg(x) = x` such that:

`f(x) = arcsin 7x => df(x) = 7/(sqrt(1 - 49x^2))`

`dg(x) = x => g(x) = x^2/2`

`int xarcsin 7x dx = (x^2arcsin 7x)/2 - (7/2)int x^2/(sqrt(1 - 49x^2)) dx`

You should use the following substitution such that:

`7x = sin t => 7dx = cos t dt`

`49x^2 = sin^2 t`

`int x^2/(sqrt(1 - 49x^2)) dx = 1/(7*49) int (sin^2 t)/(sqrt(1 - sin^2 t))cos t dt`

You should use the fundamental formula of trigonometry such that:

`cos t = sqrt(1 - sin^2 t)`

`int x^2/(sqrt(1 - 49x^2)) dx = 1/(7*49) int (sin^2 t*cos t)/cos t dt`

`int x^2/(sqrt(1 - 49x^2)) dx = 1/(7*49) int (sin^2 t) dt`

You need to use half angle identity such that:

`1/(7*49) int (sin^2 t) dt= 1/(7*49) int (1 - cos 2t)/2 dt`

`1/(7*49) int (sin^2 t) dt= 1/686(int dt - int cos 2t dt)`

`1/(7*49) int (sin^2 t) dt= t/686 - (sin 2t)/1372 + c`

`int xarcsin 7x dx = (x^2arcsin 7x)/2 - ((arcsin 7x)/392) + (sin 2(arcsin 7x))/392 + c`

You may evaluate the definite integral now such that:

`int_0^(1/7) xarcsin 7x dx = (arcsin 1)/98 - ((arcsin 1)/392) + (sin 2(arcsin 1))/392 - 0 + 0 - 0`

You need to substitute `pi/2` for `arcsin 1` such that:

`int_0^(1/7) xarcsin 7x dx = pi/196 - pi/784`

`int_0^(1/7) xarcsin 7x dx = 4pi/784 - pi/784 = 3pi/784`

**Hence, evaluating the given definite integral yields `int_0^(1/7) xarcsin 7x dx = 3pi/784.` **