Question

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

$\int x^{1/4}\sin x^{5/4}\mathrm{d}x$

Short answer

The solution of the given integral is $\int x^{1/4}\sin x^{5/4}\mathrm{d}x=-\frac{4}{5}\cos x^{5/4}+C$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int x^{1/4}\sin x^{5/4}\mathrm{d}x$

Step 2. Solving the given integral using substitution method.  

Let

$$\begin{gathered} u=x^{5/4} \\ \frac{du}{dx}=\frac{5}{4}{x}^{1/4} \\ du=\frac{5}{4}{x}^{1/4}dx \\ \frac{4}{5}du=x^{1/4}dx \end{gathered}$$

Step 3. This substitution changes the integral into  

$$\begin{gathered} \int x^{1/4}\sin x^{5/4}\mathrm{d}x=\frac{4}{5}\int \sin udu \\ \int x^{1/4}\sin x^{5/4}\mathrm{d}x=-\frac{4}{5}\cos u+C \\ \int x^{1/4}\sin x^{5/4}\mathrm{d}x=-\frac{4}{5}\cos x^{5/4}+C \end{gathered}$$