Solve ∫ left(2t^3-4tright)^3}dt

$$\int{ { \left(2 { t }^{ 3 } -4t \right) }^{ 3 } }d t$$

Evaluate

$$\frac{4t^{10}}{5}-6t^{8}+16t^{6}-16t^{4}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2t^{3}-4t\right)^{3}.

$$\int 8\left(t^{3}\right)^{3}-48\left(t^{3}\right)^{2}t+96t^{3}t^{2}-64t^{3}\mathrm{d}t$$

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$\int 8t^{9}-48\left(t^{3}\right)^{2}t+96t^{3}t^{2}-64t^{3}\mathrm{d}t$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int 8t^{9}-48t^{6}t+96t^{3}t^{2}-64t^{3}\mathrm{d}t$$

To multiply powers of the same base, add their exponents. Add 6 and 1 to get 7.

$$\int 8t^{9}-48t^{7}+96t^{3}t^{2}-64t^{3}\mathrm{d}t$$

To multiply powers of the same base, add their exponents. Add 3 and 2 to get 5.

$$\int 8t^{9}-48t^{7}+96t^{5}-64t^{3}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 8t^{9}\mathrm{d}t+\int -48t^{7}\mathrm{d}t+\int 96t^{5}\mathrm{d}t+\int -64t^{3}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$8\int t^{9}\mathrm{d}t-48\int t^{7}\mathrm{d}t+96\int t^{5}\mathrm{d}t-64\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{9}\mathrm{d}t with \frac{t^{10}}{10}. Multiply 8 times \frac{t^{10}}{10}.

$$\frac{4t^{10}}{5}-48\int t^{7}\mathrm{d}t+96\int t^{5}\mathrm{d}t-64\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{7}\mathrm{d}t with \frac{t^{8}}{8}. Multiply -48 times \frac{t^{8}}{8}.

$$\frac{4t^{10}}{5}-6t^{8}+96\int t^{5}\mathrm{d}t-64\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{5}\mathrm{d}t with \frac{t^{6}}{6}. Multiply 96 times \frac{t^{6}}{6}.

$$\frac{4t^{10}}{5}-6t^{8}+16t^{6}-64\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{3}\mathrm{d}t with \frac{t^{4}}{4}. Multiply -64 times \frac{t^{4}}{4}.

$$\frac{4t^{10}}{5}-6t^{8}+16t^{6}-16t^{4}$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$-16t^{4}+16t^{6}-6t^{8}+\frac{4t^{10}}{5}+С$$

Differentiate w.r.t. t

$$8\left(t\left(t^{2}-2\right)\right)^{3}$$