What is the integral of e^{-cos(x)}sin^3(x) ?

The integral of e^{-cos(x)}sin^3(x) is -(-2(-e^{-cos(x)}-e^{-cos(x)}cos(x))-e^{-cos(x)}sin^2(x))+C

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integral of e^{-cos(x)}sin^3(x)

\int e^{- c o s (x)} sin^{3} (x) d x

- (- 2 (- e^{- c o s (x)} - e^{- c o s (x)} cos (x)) - e^{- c o s (x)} sin^{2} (x)) + C

Solution steps

\int e^{- c o s (x)} sin^{3} (x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

= - (- e^{- u} (1 - u^{2}) - 2 (- e^{- u} u - e^{- u}))

Substitute back

u = cos (x)

= - (- e^{- c o s (x)} (1 - cos^{2} (x)) - 2 (- e^{- c o s (x)} cos (x) - e^{- c o s (x)}))

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= - (- 2 (- e^{- c o s (x)} - e^{- c o s (x)} cos (x)) - e^{- c o s (x)} sin^{2} (x))

Add a constant to the solution

= - (- 2 (- e^{- c o s (x)} - e^{- c o s (x)} cos (x)) - e^{- c o s (x)} sin^{2} (x)) + C

x \to 0 lim (cot (x))

\frac{d}{d y} (2 y - 1)

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\int \frac{1}{cos ^{2} ( x )} d x

\frac{d}{d x} (ln (x) + x)

What is the integral of e^{-cos(x)}sin^3(x) ?
The integral of e^{-cos(x)}sin^3(x) is -(-2(-e^{-cos(x)}-e^{-cos(x)}cos(x))-e^{-cos(x)}sin^2(x))+C