What is the integral from e^x to 0 of sin^3(t) ?

The integral from e^x to 0 of sin^3(t) is -1+cos(e^x)+(1-cos^3(e^x))/3

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integral from e^x to 0 of sin^3(t)

\int_{e^{x}}^{0} sin^{3} (t) d t

- 1 + cos (e^{x}) + \frac{1 - cos ^{3} ( e ^{x} )}{3}

Solution steps

\int_{e^{x}}^{0} sin^{3} (t) d t

Rewrite using trig identities

= \int_{e^{x}}^{0} (1 - cos^{2} (t)) sin (t) d t

Apply u-substitution

: \int_{c o s (e^{x})}^{1} - 1 + u^{2} d u

= \int_{c o s (e^{x})}^{1} - 1 + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{c o s (e^{x})}^{1} 1 d u + \int_{c o s (e^{x})}^{1} u^{2} d u

\int_{c o s (e^{x})}^{1} 1 d u = 1 - cos (e^{x})

\int_{c o s (e^{x})}^{1} u^{2} d u = \frac{1 - cos ^{3} ( e ^{x} )}{3}

= - (1 - cos (e^{x})) + \frac{1 - cos ^{3} ( e ^{x} )}{3}

Simplify

= - 1 + cos (e^{x}) + \frac{1 - cos ^{3} ( e ^{x} )}{3}

\int \frac{32}{x ( x ^{2} + 4 ) ^{2}} d x

y = x^{2}, y = x, 0.25, 1

\frac{x ^{2} - 3 x + 8}{x ^{2} + 9}

x^{4 x}

\frac{1}{s - 1} + \frac{1}{s + 3}

What is the integral from e^x to 0 of sin^3(t) ?
The integral from e^x to 0 of sin^3(t) is -1+cos(e^x)+(1-cos^3(e^x))/3