Question

How do you evaluate the integral of $\left( {{{\cos }^2}x\sin xdx} \right)$ from $0$ to $2\pi$ ?

Answer 1

We need to know the basic integration and differentiation formulae to solve these types of questions we need to know trigonometric table values to make the easy calculation. Also, we need to know how to apply the limit in the integral calculations. This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division.

Complete step by step answer:
The given problem is shown below,
$\int\limits_0^{2\pi } {\left( {{{\cos }^2}x\sin xdx} \right)} = ? \to \left( 1 \right)$
To solve the above equation, let’s take $u = \cos x$
So, we get $du = - \sin xdx$
(Here, the formula $\dfrac{{d\cos x}}{{dx}} = - \sin x$ )
Let’s apply the limit,
When $x = 0 \Rightarrow u = \cos 0 = 1$
When $x = 2\pi \Rightarrow u = \cos \left( {2\pi } \right) = 1$
Let’s substitute the values $u,du$ with limits $\left( {1,1} \right)$ in the equation $\left( 1 \right)$ , we get
$\left( 1 \right) \to \int\limits_0^{2\pi } {\left( {{{\cos }^2}x\sin xdx} \right)} = ?$
$\int\limits_1^1 {{u^2}} \left( { - du} \right) = - \int\limits_1^1 {{u^2}} du$
We know that
$\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
$- \int\limits_1^1 {{u^2}} du = - \left[ {\dfrac{{{u^3}}}{3}} \right]_1^1 = - \left[ {\dfrac{{{1^3}}}{3} - \dfrac{{{1^3}}}{3}} \right] = - \left[ 0 \right]$
$- \int\limits_1^1 {{u^2}} du = 0$
By substituting $u = \cos x$ and $du = - \sin xdx$ with limits $\left( {0,2\pi } \right)$ , the above equation can be modified as
$\int\limits_0^{2\pi } {\left( {{{\cos }^2}x\sin xdx} \right)} = 0$
So, the final answer is,
$\int\limits_0^{2\pi } {\left( {{{\cos }^2}x\sin xdx} \right)} = 0$

Note: Remember the basic formula of integral calculations and differential calculations. Note that if $\theta$ the value is odd $\pi$ , then the value $\cos \theta$ is equal to $- 1$ and if $\theta$ the value is even $\pi$ , then the value $\cos \theta$ is equal to $1$ . Also, note that when we replace the $x$ terms $u$ , we would apply the same condition with involvements. Because, when $x$ terms change $u$ , the limits also will change. This type of question involves the arithmetic operations like addition/subtraction/ multiplication/ division. Note that when we apply the limit we have to subtract the lower limit from the upper limit of the integral function. Also note that the denominator term would not be equal to zero, if the denominator term is zero the total value of the term is undefined or infinity.