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Question

Question: Find the integral of the given expression \[{\left( {1 - {x^2}} \right)^{0.5}}\]....

Find the integral of the given expression (1x2)0.5{\left( {1 - {x^2}} \right)^{0.5}}.

Explanation

Solution

In order to solve the given question, we will find the integrals using the substitution method. In the substitution method we will at first substitute the value of x for theta and then simplify it using formulas to obtain the required integral.

Complete step-by-step solution:
Given expression is,
(1x2)0.5\Rightarrow {\left( {1 - {x^2}} \right)^{0.5}}
Now, let x=sinθx = \sin \theta
And taking integral on the expression, we get,
(1x2)0.5dx=(1sin2θ)0.5(dsinθdθ)dθ(1)\Rightarrow \int {{{\left( {1 - {x^2}} \right)}^{0.5}}} dx = \int {{{\left( {1 - {{\sin }^2}\theta } \right)}^{0.5}}\left( {\dfrac{{d\sin \theta }}{{d\theta }}} \right)} d\theta - - - - - \left( 1 \right)
Now, according to trigonometric identity
We know that, 1sin2θ=cos2θ(2)1 - {\sin ^2}\theta = {\cos ^2}\theta - - - - - \left( 2 \right)
Thus, substituting the value of cosine function from (2) to equation (1)
We get,
(cos2θ)0.5(cosθ)dθ(3)\Rightarrow \int {{{\left( {{{\cos }^2}\theta } \right)}^{0.5}}\left( {\cos \theta } \right)} d\theta - - - - - \left( 3 \right)
Which is
cos2θdθ\Rightarrow \int {{{\cos }^2}\theta d\theta }
Now, we know that
cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1
Therefore,
cos2x=12(cos2θ1)(4)\Rightarrow {\cos ^2}x = \dfrac{1}{2}\left( {\cos 2\theta - 1} \right) - - - - - \left( 4 \right)
Thus, substituting the value of cosine function from (4) into (3)
12(cos2θ+1)dθ\Rightarrow \int {\dfrac{1}{2}\left( {\cos 2\theta + 1} \right)} d\theta
Therefore, after integrating the above function, we get,
14sin2θ+θ2+C\Rightarrow \dfrac{1}{4}\sin 2\theta + \dfrac{\theta }{2} + C
12sinθcosθ+θ2+C\Rightarrow \dfrac{1}{2}\sin \theta \cos \theta + \dfrac{\theta }{2} + C
Which is our required integral.

Additional information: In the above question we have taken use of various trigonometric identities and several standard trigonometric integrals. The trigonometric identities that we have used in the above question are as follows:
1sin2θ=cos2θ1 - {\sin ^2}\theta = {\cos ^2}\theta
cos2x=2cos2x1\cos 2x = 2{\cos ^2}x - 1
cos2x=1sin2x\cos 2x = 1 - {\sin ^2}x
sin2x=2sinxcosx\sin 2x = 2\sin x\cos x
The standard trigonometric functions that we have taken use of in solving the above question is:
sinxdx=cosx\int {\sin xdx = \cos x}
It is important to keep in mind all the trigonometric identities and standard integrals in order to solve integration.

Note: We have solved the given question by the substitution method in which we substitute the value of x to t by substituting x=g (t). Other than this method, there are two other methods for solving integration, they are, Integration by partial fraction method and integration by parts. Some integration solution might involve using techniques of two methods