Evaluate $\displaystyle \int{\sin^2{x}}\,dx$

The integral of the sine square of angle $x$ should be evaluated with respect to $x$ in this indefinite integration math problem.

The trigonometric function in this indefinite integral problem is in square form and there is no integration formula to find the integral of the square of sine function but there are trigonometric integration rules to find the integration of the trigonometric functions in calculus. So, it is better to reduce the power of the sine function by using the power reduction trigonometric identities.

Reduce the power of the sine square function

Let’s use the power reduction identity of sine function to reduce the power of the sine function. Now, the square of sine function is successfully converted as a fraction in terms of cosine.

$\implies$ $\displaystyle \int{\sin^2{x}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1-\cos{2x}}{2}}\,dx$

Separate the constant factor from Integration

The trigonometric function in square form is successfully converted as a trigonometric function in fraction form. There is a trigonometric expression in the numerator but the denominator is a number, which is a constant and it expresses that there is a constant factor is involved in the function. So, it is better to separate the constant factor from the function and it allows us to find the integral easily.

Each expression in both numerator and denominator can be written mathematically as a product of two factors.

$\implies$ $\displaystyle \int{\dfrac{1-\cos{2x}}{2}}\,dx$ $=\,\,$ $\displaystyle \int{\dfrac{1 \times (1-\cos{2x})}{2 \times 1}}\,dx$

It seems the fraction is formed by the product of two fractions. So, the fraction can be split as a product of two fraction as per the principle of the multiplication of the fractions.

$=\,\,$ $\displaystyle \int{\dfrac{1}{2} \times \dfrac{(1-\cos{2x})}{1}}\,dx$

$=\,\,$ $\displaystyle \int{\dfrac{1}{2} \times (1-\cos{2x})}\,dx$

The fraction one half is a coefficient of the function one minus cosine of two times angle $x$ and it is a constant factor. So, it can be taken out from the integral operation as per the constant multiple rule of the integration.

$=\,\,$ $\dfrac{1}{2} \times \displaystyle \int{(1-\cos{2x})}\,dx$

Find the Indefinite integral of the function

The function one minus cosine of double angle expresses the difference between two functions, and its integration can be evaluated by finding the difference of their integrals as per the subtraction rule of the integration.

$=\,\,$ $\dfrac{1}{2} \times \bigg(\displaystyle \int{1}\,dx-\displaystyle \int{\cos{2x}}\,dx\bigg)$

Firstly, let us find the integration of the one with respect to $x$. The integral of one with respect to $x$ is equal to $x$ plus the integral constant as per the integral rule of one.

$=\,\,$ $\dfrac{1}{2} \times \bigg(x+c_1-\displaystyle \int{\cos{2x}}\,dx\bigg)$

Now, let’s find the integration of second term in the second factor of the expression. There is an integration rule to find the integral of the cosine function but the double angle is involved in the cosine function. That is why, it is not possible to use the integral rule of cosine function.

However, Integration formula of cosine function can be used by representing the double angle by a variable. So, let’s denote the double angle $2x$ by a variable $y$.

$y \,=\, 2x$

Now, let’s differentiate the equation with respect to $x$ to convert the integral of the function in terms of $x$ into the integral of the function in terms of $y$.

$\implies$ $\dfrac{d}{dx}{(y)} \,=\, \dfrac{d}{dx}{(2x)}$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}{(2 \times x)}$

$\implies$ $\dfrac{dy}{dx} \,=\, 2 \times \dfrac{d}{dx}{x}$

$\implies$ $\dfrac{dy}{dx} \,=\, 2 \times \dfrac{dx}{dx}$

$\implies$ $\dfrac{dy}{dx} \,=\, 2 \times 1$

$\implies$ $\dfrac{dy}{dx} \,=\, 2$

$\implies$ $dy \,=\, 2 \times dx$

$\implies$ $\dfrac{dy}{2} \,=\, \dfrac{2 \times dx}{2}$

$\implies$ $\dfrac{dy}{2} \,=\, \dfrac{\cancel{2} \times dx}{\cancel{2}}$

$\implies$ $\dfrac{dy}{2} \,=\, dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx \,=\, \dfrac{dy}{2}$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\cos{y}}\,\times\,\bigg(\dfrac{dy}{2}\bigg)\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\cos{y}}\,\times\,\bigg(\dfrac{1 \times dy}{2 \times 1}\bigg)\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\cos{y}}\,\times\,\bigg(\dfrac{1}{2} \times \dfrac{dy}{1}\bigg)\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\cos{y}}\,\times\,\bigg(\dfrac{1}{2} \times dy\bigg)\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\cos{y}}\,\times\,\dfrac{1}{2} \times dy\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\dfrac{1}{2} \times \cos{y}}\,\times\,dy\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\displaystyle \int{\dfrac{1}{2} \times \cos{y}}\,dy\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\dfrac{1}{2} \times \displaystyle \int{\cos{y}}\,dy\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \Bigg(x+c_1-\dfrac{1}{2}\displaystyle \int{\cos{y}}\,dy\Bigg)$

$=\,\,$ $\dfrac{1}{2} \times \bigg(x+c_1-\dfrac{1}{2} \times (\sin{y}+c_2)\bigg)$

$=\,\,$ $\dfrac{1}{2} \times \bigg((x+c_1)-\dfrac{1}{2} \times (\sin{y}+c_2)\bigg)$

$=\,\,$ $\dfrac{1}{2} \times (x+c_1)$ $-$ $\dfrac{1}{2} \times \dfrac{1}{2} \times (\sin{y}+c_2)$

$=\,\,$ $\dfrac{1 \times (x+c_1)}{2}$ $-$ $\dfrac{1 \times 1 \times (\sin{y}+c_2)}{2 \times 2}$

$=\,\,$ $\dfrac{1 \times (x+c_1)}{2}$ $-$ $\dfrac{2 \times (\sin{y}+c_2)}{2}$

$=\,\,$ $\dfrac{1 \times (x+c_1)}{2}$ $-$ $\dfrac{\cancel{2} \times (\sin{y}+c_2)}{\cancel{2}}$

$=\,\,$ $\dfrac{x+c_1}{2}$ $-$ $(\sin{y}+c_2)$

$=\,\,$ $\dfrac{x}{2}$ $+$ $\dfrac{c_1}{2}$ $-$ $\sin{y}$ $-$ $c_2$

$=\,\,$ $\dfrac{x}{2}$ $-$ $\sin{y}$ $+$ $\dfrac{c_1}{2}$ $-$ $c_2$

$=\,\,$ $\dfrac{x}{2}$ $-$ $\sin{y}$ $+$ $c$

$=\,\,$ $\dfrac{x}{2}$ $-$ $\sin{2x}$ $+$ $c$

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