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

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

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

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

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

Solution steps

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

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 (t)

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

Use the Pythagorean identity:

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

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

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

Add a constant to the solution

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

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