What is the integral from 1 to sin(x) of 7t^6 ?

The integral from 1 to sin(x) of 7t^6 is sin^7(x)-1

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integral from 1 to sin(x) of 7t^6

\int_{1}^{s i n (x)} 7 t^{6} d t

sin^{7} (x) - 1

Solution steps

\int_{1}^{s i n (x)} 7 t^{6} d t

Take the constant out:

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

= 7 \cdot \int_{1}^{s i n (x)} t^{6} d t

Apply the Power Rule

: [\frac{t ^{7}}{7}]_{1}^{s i n (x)}

= 7 [\frac{t ^{7}}{7}]_{1}^{s i n (x)}

Compute the boundaries:

\frac{sin ^{7} ( x )}{7} - \frac{1}{7}

= 7 (\frac{sin ^{7} ( x )}{7} - \frac{1}{7})

Simplify

= sin^{7} (x) - 1

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What is the integral from 1 to sin(x) of 7t^6 ?
The integral from 1 to sin(x) of 7t^6 is sin^7(x)-1