Question

Integrate $\int {\sqrt {1 - {t^2}} dt}$

Answer 1

Here, we will assume the variable of the given function to be some trigonometric function. We will then differentiate the assumed value and substitute it in the given function. Then by using the trigonometric identity, we will simplify the integrand. Then by using the concept of integration, we integrate the function. Integration is the process of adding the small parts to find the whole parts.

Formula Used:
We will use the following formula:

1. Differentiation formula: $\dfrac{d}{{d\theta }}\left( {\sin \theta } \right) = \cos \theta$
2. Trigonometric Identity: $1 - {\sin ^2}\theta = {\cos ^2}\theta$
3. Trigonometric Identity: ${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$
4. Trigonometric Identity: $\sin 2\theta = 2\sin \theta \cos \theta$
5. Integration formula: $\int {cdx = xc + C}$
6. Integration formula: $\int {\cos 2xdx = \dfrac{{\sin 2x}}{2} + C}$

Complete step by step solution:
We are given that $\int {\sqrt {1 - {t^2}} dt}$
Let $t = \sin \theta$
Now, we will differentiate the variable $t$ with respect to the variable $\theta$ using the formula $\dfrac{d}{{d\theta }}\left( {\sin \theta } \right) = \cos \theta$.
$\Rightarrow \dfrac{{dt}}{{d\theta }} = \cos \theta$
By rewriting the equation, we get
$\Rightarrow dt = \cos \theta \cdot d\theta$
By substituting the variable and the differentiation of the variable in $\int {\sqrt {1 - {t^2}} dt}$, we get
$\int {\sqrt {1 - {t^2}} dt} = \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta$
By using the trigonometric identity $1 - {\sin ^2}\theta = {\cos ^2}\theta$, we get
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta = \int {\sqrt {{{\cos }^2}\theta } } \cdot \cos \theta d\theta$
We know that the square of a square root of a number is the number. So, we get
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta = \int {\cos \theta } \cdot \cos \theta d\theta$
By multiplying the terms, we get
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta = \int {{{\cos }^2}\theta } d\theta$
Using the trigonometric identity ${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$, we get
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta = \int {\dfrac{{1 + \cos 2\theta }}{2}} d\theta$
By separating the terms, we get
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta = \int {\dfrac{1}{2} + \dfrac{{\cos 2\theta }}{2}} d\theta$
$\Rightarrow \int {\sqrt {1 - {{\sin }^2}\theta } } \cdot \cos \theta d\theta = \int {\dfrac{1}{2} + \dfrac{{\cos 2\theta }}{2}} d\theta$
Integrating the terms using the formula $\int {cdx = xc + C}$ and $\int {\cos 2xdx = \dfrac{{\sin 2x}}{2} + C}$, we get
$\Rightarrow \int {\dfrac{1}{2} + \dfrac{{\cos 2\theta }}{2}} d\theta = \dfrac{1}{2}\theta + \dfrac{1}{2} \cdot \dfrac{{\sin 2\theta }}{2} + C$
Using the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$, we get
$\Rightarrow \int {\dfrac{1}{2} + \dfrac{{\cos 2\theta }}{2}} d\theta = \dfrac{1}{2}\theta + \dfrac{1}{2} \cdot \dfrac{{2\sin \theta \cos \theta }}{2} + C$
By rewriting the variable $\theta$ in terms of the variable $t$, we get
$\Rightarrow \int {\dfrac{1}{2} + \dfrac{{\cos 2\theta }}{2}} d\theta = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}t} \right) + \dfrac{1}{2} \cdot t \cdot \sqrt {1 - {t^2}} + C$

Therefore, the integration of $\int {\sqrt {1 - {t^2}} dt}$ is $\dfrac{1}{2}\left( {{{\sin }^{ - 1}}t} \right) + \dfrac{1}{2} \cdot t \cdot \sqrt {1 - {t^2}} + C$.

Note:
We should always be conscious that whenever we are using the method of substitution in integration or whenever we are given the limits we should always change the upper limit and lower limit according to the substitution. The variable to be integrated also changes according to the substitution. When the integrand is in trigonometric function, then it satisfies the basic properties of integration. The given integral function is an indefinite integral since there are no limits in the integral.