Q. 74 - Calculus

Question

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

${\int }_0^{\sqrt{4x}}x\sin x^{2}dx$

Step-by-Step Solution

Answer

The solution of the given integral is ${\int }_0^{\sqrt{4x}}x\sin x^{2}dx=-\frac{1}{2}[\cos 4x-1]$.

Step 1. Given Information

Solving the given integrals.

${\int }_0^{\sqrt{4x}}x\sin x^{2}dx$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=x^2 \\ \frac{du}{dx}=2x \\ du=2xdx \\ \frac{1}{2}du=xdx \end{gathered}$$

Step 3. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_0^{\sqrt{4x}}x\sin x^{2}dx=\frac{1}{2}{\int }_{x=0}^{x=\sqrt{4x}}\sin udu \\ {\int }_0^{\sqrt{4x}}x\sin x^{2}dx=\frac{1}{2}{[-\cos u]}_{x=0}^{x=\sqrt{4x}} \\ {\int }_0^{\sqrt{4x}}x\sin x^{2}dx=-\frac{1}{2}{\left[\cos u\right]}_{x=0}^{x=\sqrt{4x}} \\ {\int }_0^{\sqrt{4x}}x\sin x^{2}dx=-\frac{1}{2}{\left[\cos x^2\right]}_{x=0}^{x=\sqrt{4x}} \\ {\int }_0^{\sqrt{4x}}x\sin x^{2}dx=-\frac{1}{2}[\cos {\left(\sqrt{4x}\right)}^2-\cos 0] \\ {\int }_0^{\sqrt{4x}}x\sin x^{2}dx=-\frac{1}{2}[\cos 4x-1] \end{gathered}$$