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Question: How do you integrate \( \int {\sqrt {1 - {x^2}} dx} \) by trigonometric substitution?...

How do you integrate ∫1−x2dx\int {\sqrt {1 - {x^2}} dx} by trigonometric substitution?

Explanation

Solution

Hint : The operation of integration, up to an additive constant, is the inverse of the operation of differentiation. For this reason, the term integral may also refer to the related notion of the antiderivative, called an indefinite integral, a function FF whose derivative is the given function ff
In this case, it is written:
F(x)=∫f(x)dxF(x) = \int {f(x)dx}

Complete step-by-step answer :
As the integrated function is defined for x∈[−1,1]x \in \left[ { - 1,1} \right] , you can substitute:
x=sin⁡tx = \sin t with t∈[−π2,π2]t \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]
dx=cos⁡tdtdx = \cos tdt
So, the integral becomes:
∫1−x2dx=∫1−sin⁡2tcos⁡tdt=∫cos⁡2tcos⁡tdt\int {\sqrt {1 - {x^2}} dx = \int {\sqrt {1 - {{\sin }^2}t} \cos tdt = \int {\sqrt {{{\cos }^2}t} \cos tdt} } }
In the given interval cos⁡t\cos t is positive, so cos⁡2t=cos⁡t\sqrt {{{\cos }^2}t} = \cos t :
∫1−x2dx=∫cos⁡2tdt\int {\sqrt {1 - {x^2}} dx = \int {{{\cos }^2}tdt} }
Now we can use the identity:
cos⁡2t=1+cos⁡(2t)2 ∫1−x2dx=∫1+cos⁡(2t)2dt=∫dt2+14  {\cos ^2}t = \dfrac{{1 + \cos (2t)}}{2} \\\ \int {\sqrt {1 - {x^2}} dx = \int {\dfrac{{1 + \cos (2t)}}{2}} } dt = \int {\dfrac{{dt}}{2} + \dfrac{1}{4}} \\\
∫cos⁡(2t)d(2t)=12t+14sin⁡2t=12(t+sin⁡tcos⁡t)\int {\cos (2t)d(2t) = \dfrac{1}{2}t + \dfrac{1}{4}\sin 2t = \dfrac{1}{2}(t + \sin t\cos t)}
To substitute back xx we note that:
x=sin⁡tx = \sin t for
t∈[−π2,π2]⇒t=arcsin⁡x cos⁡t=1−sin⁡2t=1−x2  t \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \Rightarrow t = \arcsin x \\\ \cos t = \sqrt {1 - {{\sin }^2}t} = \sqrt {1 - {x^2}} \\\
Finally:
∫1−x2dx=12(arcsin⁡x+x1−x2)+C\int {\sqrt {1 - {x^2}} dx = \dfrac{1}{2}(\arcsin x + x\sqrt {1 - {x^2}} )} + C\dfrac{{}}{{}}
So, the correct answer is “ ∫1−x2dx=12(arcsin⁡x+x1−x2)+C\int {\sqrt {1 - {x^2}} dx = \dfrac{1}{2}(\arcsin x + x\sqrt {1 - {x^2}} )} + C\dfrac{{}}{{}} ”.

Note : Integration is a method of adding values on a large scale, where we cannot perform general addition operation. But there are multiple methods of integration, which are used in mathematics to integrate the functions (which is easier to evaluate the original integral). The different methods of integration include:
I.Integration by substitution
II.Integration by parts
III.Integration using trigonometric identities
IV.Integration of some particular function
V.Integration by partial fraction