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

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

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integral from a to x^3 of sin^3(t)

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

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

Solution steps

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

Rewrite using trig identities

= \int_{a}^{x^{3}} (1 - cos^{2} (t)) sin (t) d t

Apply u-substitution

: \int_{c o s (a)}^{c o s (x^{3})} - 1 + u^{2} d u

= \int_{c o s (a)}^{c o s (x^{3})} - 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 (a)}^{c o s (x^{3})} 1 d u + \int_{c o s (a)}^{c o s (x^{3})} u^{2} d u

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

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

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

Simplify

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

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