Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int x^3\cos x^4\mathrm{d}x$

Short answer

The solution of the given integral is $\int x^3\cos x^4\mathrm{d}x=\frac{1}{4}\sin x^4+C$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int x^3\cos x^4\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

$$\begin{gathered} u=x^4 \\ \frac{du}{dx}=4x^3 \\ du=4x^{3}dx \\ \frac{1}{4}du=x^{3}dx \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int x^3\cos x^4\mathrm{d}x=\frac{1}{4}\int \cos u\mathrm{du} \\ \int x^3\cos x^4\mathrm{d}x=\frac{1}{4}\sin u+C \\ \int x^3\cos x^4\mathrm{d}x=\frac{1}{4}\sin x^4+C \end{gathered}$$