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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use integration by parts such that:

`int udv = uv - int vdu`

Considering `u = arcsin 7x`  and `dv = xdx`  yields:

`u = arcsin 7x => du = 7/(sqrt(1 - x^2))`

`dv = xdx => v = x^2/2`

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

You should use the following trigonometric substitution such that:

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

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

You should use the fundamental formula of trigonometry such that:

`1 - sin^2 t = cos^2 t`

`int (sin^2 t)/(sqrt(1 - sin^2 t)) cost dt = int (sin^2 t)/(sqrt(cos^2 t)) cost dt`

`int (sin^2 t)/(cos t) cost dt = int sin^2 t dt`

You need to use the following trigonometric identity such that:

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

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

Using the property of linearity yields:

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

`int sin^2 t dt = (1/2)t - (sin 2t)/4 + c`

Substituting back arcsin x for t yields:

`(7/2) int x^2/sqrt(1 - x^2)= (7/4)arcsin x- 7(sin 2arcsin x)/8 + c`

`int x arcsin(7x) dx = (x^2/2)arcsin 7x - (7/4)arcsin x+ 7(sin 2arcsin x)/8 + c`

`int_0^(1/7) x arcsin(7x) dx = (1/98)arcsin1 - (7/4)arcsin1/7 + 7(sin 2arcsin (1/7))/8 - (1/98)arcsin0+ (7/4)arcsin0- 7(sin 2arcsin (0))/8`

`int_0^(1/7) x arcsin(7x) dx = pi/196 - (7/4)arcsin 1/7 + 7(sin 2arcsin (1/7))/8`

Hence, evaluating the given definite integral yields `int_0^(1/7) x arcsin(7x) dx = pi/196 - (7/4)arcsin 1/7 + 7(sin 2arcsin (1/7))/8.`

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