Solve ∫ t^3operatorname{ch}tdt

$$\int t ^ { 3 } \operatorname { ch } t d t$$

Evaluate

$$\frac{cht^{5}}{5}+С$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$\int t^{4}ch\mathrm{d}t$$

Factor out the constant using \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

$$ch\int t^{4}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{4}\mathrm{d}t with \frac{t^{5}}{5}.

$$ch\times \frac{t^{5}}{5}$$

Simplify.

$$\frac{cht^{5}}{5}$$

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.

$$\frac{cht^{5}}{5}+С$$

Differentiate w.r.t. t

$$cht^{4}$$