What is the integral from 1 to sin(x) of 4t^3 ?

The integral from 1 to sin(x) of 4t^3 is sin^4(x)-1

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integral from 1 to sin(x) of 4t^3

\int_{1}^{s i n (x)} 4 t^{3} d t

sin^{4} (x) - 1

Solution steps

\int_{1}^{s i n (x)} 4 t^{3} d t

Take the constant out:

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

= 4 \cdot \int_{1}^{s i n (x)} t^{3} d t

Apply the Power Rule

: [\frac{t ^{4}}{4}]_{1}^{s i n (x)}

= 4 [\frac{t ^{4}}{4}]_{1}^{s i n (x)}

Compute the boundaries:

\frac{sin ^{4} ( x )}{4} - \frac{1}{4}

= 4 (\frac{sin ^{4} ( x )}{4} - \frac{1}{4})

Simplify

= sin^{4} (x) - 1

\int_{0}^{\frac{1}{4}} sin^{2} (π x) d x

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\int_{2}^{4} x^{2} - 4 x + 4 d x

\int_{1}^{x} 1 d x

What is the integral from 1 to sin(x) of 4t^3 ?
The integral from 1 to sin(x) of 4t^3 is sin^4(x)-1