Q. 55 - 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.)

$\int \frac{x\cos x^2}{\sqrt{\sin x^2}}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\sqrt{{\mathrm{sinx}}^2}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals. 

$\int \frac{x\cos x^2}{\sqrt{\sin x^2}}\mathrm{d}x$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u={\mathrm{sinx}}^2 \\ \frac{du}{dx}=\frac{\mathrm{d}}{\mathrm{d}x}{\mathrm{sinx}}^2 \\ \frac{du}{dx}=2x\cos x^2 \\ du=2x\cos x^{2}dx \\ \frac{1}{2}du=x\cos x^{2}dx \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \frac{x\cos x^2}{\sqrt{\sin x^2}}\mathrm{d}x=\frac{1}{2}\int \frac{1}{\sqrt{u}}\mathrm{du} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\frac{1}{2}\int \frac{1}{{\mathrm{u}}^{1/2}}\mathrm{du} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\frac{1}{2}\int {\mathrm{u}}^{-1/2}\mathrm{du} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\frac{1}{2}\left[\frac{{\mathrm{u}}^{-1/2+1}}{-1/2+1}\right]+\mathrm{C} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\frac{1}{2}\left[\frac{{\mathrm{u}}^{1/2}}{1/2}\right]+\mathrm{C} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\frac{1}{2}·2\sqrt{\mathrm{u}}+\mathrm{C} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\sqrt{\mathrm{u}}+\mathrm{C} \\ \int \frac{{\mathrm{xcosx}}^2}{\sqrt{{\mathrm{sinx}}^2}}\mathrm{dx}=\sqrt{{\mathrm{sinx}}^2}+\mathrm{C} \end{gathered}$$