What is the integral from 1 to sin(x) of 13t^{12} ?

The integral from 1 to sin(x) of 13t^{12} is sin^{13}(x)-1

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integral from 1 to sin(x) of 13t^{12}

\int_{1}^{s i n (x)} 13 t^{12} d t

sin^{13} (x) - 1

Solution steps

\int_{1}^{s i n (x)} 13 t^{12} d t

Take the constant out:

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

= 13 \cdot \int_{1}^{s i n (x)} t^{12} d t

Apply the Power Rule

: [\frac{t ^{13}}{13}]_{1}^{s i n (x)}

= 13 [\frac{t ^{13}}{13}]_{1}^{s i n (x)}

Compute the boundaries:

\frac{sin ^{13} ( x )}{13} - \frac{1}{13}

= 13 (\frac{sin ^{13} ( x )}{13} - \frac{1}{13})

Simplify

= sin^{13} (x) - 1

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What is the integral from 1 to sin(x) of 13t^{12} ?
The integral from 1 to sin(x) of 13t^{12} is sin^{13}(x)-1