Q. 31 - 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.) 

$\int \sin \pi x\cos \pi x\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \mathrm{sin\pi xcos\pi xdx}=-\frac{1}{2\mathrm{\pi }}{\cos }^2\mathrm{\pi x}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals. 

$\int \sin \pi x\cos \pi x\mathrm{d}x$

Step 2. Solving the given integral using substitution method.

Let 

$$\begin{gathered} u=\cos \pi x \\ \frac{du}{dx}=-\mathrm{\pi sin\pi x} \\ d\mathrm{u}=-\mathrm{\pi sin\pi x}d\mathrm{x} \\ -\frac{1}{\mathrm{\pi }}d\mathrm{u}=\mathrm{sin\pi x}d\mathrm{x} \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \sin \pi x\cos \pi x\mathrm{d}x=-\frac{1}{\mathrm{\pi }}\int u\mathrm{du} \\ \int \mathrm{sin\pi xcos\pi xdx}=-\frac{1}{\mathrm{\pi }}\left[\frac{{\mathrm{u}}^{1+1}}{1+1}\right]+\mathrm{C} \\ \int \mathrm{sin\pi xcos\pi xdx}=-\frac{1}{\mathrm{\pi }}\left[\frac{{\mathrm{u}}^2}{2}\right]+\mathrm{C} \\ \int \mathrm{sin\pi xcos\pi xdx}=-\frac{1}{\mathrm{\pi }}\left[\frac{{\cos }^2\mathrm{\pi x}}{2}\right]+\mathrm{C} \\ \int \mathrm{sin\pi xcos\pi xdx}=-\frac{1}{2\mathrm{\pi }}{\cos }^2\mathrm{\pi x}+\mathrm{C} \end{gathered}$$