What is the integral from 6 to sin(x) of x^3 ?

The integral from 6 to sin(x) of x^3 is (sin^4(x))/4-324

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integral from 6 to sin(x) of x^3

\int_{6}^{s i n (x)} x^{3} d x

\frac{sin ^{4} ( x )}{4} - 324

Solution steps

\int_{6}^{s i n (x)} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{6}^{s i n (x)}

= [\frac{x ^{4}}{4}]_{6}^{s i n (x)}

Compute the boundaries:

\frac{sin ^{4} ( x )}{4} - 324

= \frac{sin ^{4} ( x )}{4} - 324

\int_{0}^{4} (x - 6)^{2} d x

\int_{x}^{y} ln (sin (t)) d t

\int_{- π}^{π} cos (t) d t

\int_{0}^{2} e^{y^{3}} y^{2} d y

\int_{0}^{1} x^{3} \cdot x^{2} + 1 d x

What is the integral from 6 to sin(x) of x^3 ?
The integral from 6 to sin(x) of x^3 is (sin^4(x))/4-324