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Popular >

integral of xe^{arcsin(x)}

Solution

\int x e^{a r c s i n (x)} d x

Solution

- \frac{e ^{a r c s i n (x)} cos ( 2 arcsin ( x ) )}{5} + \frac{e ^{a r c s i n (x)} sin ( 2 arcsin ( x ) )}{10} + C

Solution steps

\int x e^{a r c s i n (x)} d x

Apply u-substitution

: \int e^{u} cos (u) sin (u) d u

= \int e^{u} cos (u) sin (u) d u

Rewrite using trig identities

= \int \frac{1}{2} e^{u} sin (2 u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{u} sin (2 u) d u

Apply Integration By Parts

: - \frac{1}{2} e^{u} cos (2 u) - \int - \frac{1}{2} e^{u} cos (2 u) d u

= \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - \int - \frac{1}{2} e^{u} cos (2 u) d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - (- \frac{1}{2} \cdot \int e^{u} cos (2 u) d u))

Apply Integration By Parts

: e^{u} \frac{1}{2} sin (2 u) - \int e^{u} \frac{1}{2} sin (2 u) d u

= \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - (- \frac{1}{2} (e^{u} \frac{1}{2} sin (2 u) - \int e^{u} \frac{1}{2} sin (2 u) d u)))

Therefore

\int e^{u} \frac{1}{2} sin (2 u) d u = \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - (- \frac{1}{2} (e^{u} \frac{1}{2} sin (2 u) - \int e^{u} \frac{1}{2} sin (2 u) d u)))

Isolate

\int e^{u} \frac{1}{2} sin (2 u) d u

= - \frac{e ^{u} cos ( 2 u )}{5} + \frac{e ^{u} sin ( 2 u )}{10}

Substitute back

u = arcsin (x)

= - \frac{e ^{a r c s i n (x)} cos ( 2 arcsin ( x ) )}{5} + \frac{e ^{a r c s i n (x)} sin ( 2 arcsin ( x ) )}{10}

Add a constant to the solution

= - \frac{e ^{a r c s i n (x)} cos ( 2 arcsin ( x ) )}{5} + \frac{e ^{a r c s i n (x)} sin ( 2 arcsin ( x ) )}{10} + C

Graph

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