Question

How do you find the integral $\int {(\sin x)(\cos x)dx}$ ?

Answer 1

Here, we need to find the value of $\int {(\sin x)(\cos x)dx}$ . There are many ways to find the answer. One should know some trigonometry formulas such as ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos 2x = 2{\cos ^2}x - 1$ . To find integration of sin x cos x, we will use the formula $\sin 2x = 2\sin x\cos x$ . By using this, and evaluating this, we will get the final output.

Complete step by step answer:
Given that, $\int {(\sin x)(\cos x)dx}$. We have many methods to solve this problem. We will use all of them one by one to solve them.The different methods used are as below:
First we will substitute $u = \sin x$ and applying this, we will get,
Let,
$u = \sin x$
$\Rightarrow du = \cos xdx$
For,
$\int {(\sin x)(\cos x)dx}$
Substitute the value of u and du in the above integral, we will get,
$\Rightarrow \int {(u)(du)}$
Removing the brackets, we will get,
$\Rightarrow \int {udu}$
$\Rightarrow \dfrac{{{u^2}}}{2} + C$ where C is the constant of integration

Again, we will substitute the value of u, we will get,
$\Rightarrow \dfrac{{{{(\sin x)}^2}}}{2} + C$
$\Rightarrow \dfrac{{{{\sin }^2}x}}{2} + C$
We will use the identity here as below:
{\sin ^2}x + {\cos ^2}x = 1$$$$ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x
So applying this, we will get,
$\Rightarrow \dfrac{{1 - {{\cos }^2}x}}{2} + C$
$\Rightarrow \dfrac{1}{2} + \dfrac{{ - {{\cos }^2}x}}{2} + C$
$\Rightarrow - \dfrac{{{{\cos }^2}x}}{2} + \dfrac{1}{2} + C$
Here, $\dfrac{1}{2}$ is absorbed into C as C represents any constant, we will get,
$\Rightarrow - \dfrac{{{{\cos }^2}x}}{2} + C$

First we will substitute $u = \sin x$ and applying this, we will get,
Let,
$u = \cos x$
$\Rightarrow du = - \sin xdx$
For,
$\int {(\sin x)(\cos x)dx}$
$\Rightarrow\int {(\cos x)(\sin x)dx}$
Substitute the value of u and du in the above integral, we will get,
$\Rightarrow \int {(u)( - du)}$
Removing the brackets, we will get,
$\Rightarrow - \int {udu}$
$\Rightarrow - \dfrac{{{u^2}}}{2} + C$
Again, we will substitute the value of u, we will get,
$\Rightarrow - \dfrac{{{{(\cos x)}^2}}}{2} + C$
$\Rightarrow - \dfrac{{{{\cos }^2}x}}{2} + C$

Here, we will use the formula of sin2x as below:
$\sin 2x = 2\sin x\cos x$
$\Rightarrow \dfrac{{\sin 2x}}{2} = \sin x\cos x$
$\Rightarrow \int {(\sin x)(\cos x)dx}$
Substitute the value, we will get,
$\Rightarrow \int {\dfrac{{\sin 2x}}{2}dx}$
$\Rightarrow \dfrac{1}{2}\int {\sin 2xdx}$
Let, $u = 2x$
$\Rightarrow du = 2dx$
$\Rightarrow dx = \dfrac{{du}}{2}$

Now, we will substitute the values of 2x and dx, we will get,
$\Rightarrow \dfrac{1}{2}\int {\sin u\dfrac{{du}}{2}}$
$\Rightarrow \dfrac{1}{4}\int {\sin udu}$
We know that, $\int {\sin xdx = - \cos x + C}$ and so applying this, we will get,
$\Rightarrow \dfrac{1}{4}( - \cos u) + C$
$\Rightarrow - \dfrac{1}{4}\cos u + C$
$\Rightarrow - \dfrac{1}{4}\cos 2x + C$
We will use the formula $\cos 2x = 2{\cos ^2}x - 1$ and so using this, we will get,
$\Rightarrow - \dfrac{1}{4}\left( {2{{\cos }^2}x - 1} \right) + C$
Removing the brackets, we will get,
$\Rightarrow \dfrac{{ - {{\cos }^2}x}}{2} + \dfrac{1}{4} + C$
On simplifying this, we will get,
$\Rightarrow \dfrac{{ - {{\cos }^2}x}}{2} + C$

Hence, $\int {(\sin x)(\cos x)dx} = \dfrac{{ - {{\cos }^2}x}}{2} + C$.

Note: Integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles.