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Question: What is the Integration of \(\left[ {\dfrac{{x + 3}}{{\sqrt {9 - {x^2}} }}} \right]\)?...

What is the Integration of [x+39−x2]\left[ {\dfrac{{x + 3}}{{\sqrt {9 - {x^2}} }}} \right]?

Explanation

Solution

In this question, first we have to split both the terms in the numerator and then we will find the integration of both of them separately. In the first integral with the numerator as xx, we will use the substitution method and in the second integral we can easily find it as an integral of sin⁡−1x{\sin ^{ - 1}}x format.

Complete step by step answer:
In the above question, we have to find the integral of [x+39−x2]\left[ {\dfrac{{x + 3}}{{\sqrt {9 - {x^2}} }}} \right]. Let the integral be equal to I.
I=∫x+39−x2dxI = \int {\dfrac{{x + 3}}{{\sqrt {9 - {x^2}} }}dx}
Now split the numerator in two terms.
⇒I=∫x9−x2dx+∫39−x2dx\Rightarrow I = \int {\dfrac{x}{{\sqrt {9 - {x^2}} }}dx} + \int {\dfrac{3}{{\sqrt {9 - {x^2}} }}dx}
Let I1=∫x9−x2dx{I_1} = \int {\dfrac{x}{{\sqrt {9 - {x^2}} }}dx} and I2=∫39−x2dx{I_2} = \int {\dfrac{3}{{\sqrt {9 - {x^2}} }}dx}
⇒I=I1+I2................(1)\Rightarrow I = {I_1} + {I_2}................\left( 1 \right)

Now let’s calculate I1{I_1}
I1=∫x9−x2dx{I_1} = \int {\dfrac{x}{{\sqrt {9 - {x^2}} }}dx}
We will use substitution method here
Let 9−x2=u\sqrt {9 - {x^2}} = u
Squaring both sides, we get 9−x2=u29 - {x^2} = {u^2}
Now transposing the terms, we get x2=9−u2{x^2} = 9 - {u^2}
Now, on differentiating both sides
2xdx=−2udu2xdx = - 2udu
We know that the differentiation of constant term is zero.
Therefore, on dividing both sides by 22, we get
xdx=−uduxdx = - udu

Now substituting the value of xdx and 9−x2xdx\,and\,\sqrt {9 - {x^2}} in the integral.
⇒I1=∫−uduu\Rightarrow {I_1} = \int {\dfrac{{ - udu}}{u}}
On division, we get
⇒I1=∫−1du\Rightarrow {I_1} = \int { - 1du}
We know that the integration of 1  is  u1\,\,is\,\,u.
⇒I1=−u+c1\Rightarrow {I_1} = - u + {c_1}
Now substitute the value of u in the above equation.
⇒I1=−9−x2+c1\Rightarrow {I_1} = - \sqrt {9 - {x^2}} + {c_1}
Here c1{c_1} is the constant of integration.
Now we will evaluate I2{I_2}.
I2=∫39−x2dx{I_2} = \int {\dfrac{3}{{\sqrt {9 - {x^2}} }}dx}

We can also write it as
⇒I2=3∫1(3)2−x2dx\Rightarrow {I_2} = 3\int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {x^2}} }}dx}
We know that ∫1a2−x2dx=sin⁡−1xa\int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}dx = {{\sin }^{ - 1}}\dfrac{x}{a}}
Therefore, on substituting the value a=3a = 3.
⇒I2=3sin⁡−1x3+c2\Rightarrow {I_2} = 3{\sin ^{ - 1}}\dfrac{x}{3} + {c_2}
Now substitute the value of I1  and  I2{I_1}\,\,and\,\,{I_2} in equation (1)\left( 1 \right).
⇒I=−9−x2 +c1+3sin⁡−1x3+c2\Rightarrow I = - \sqrt {9 - {x^2}} \, + {c_1} + 3{\sin ^{ - 1}}\dfrac{x}{3} + {c_2}
Now, put c=c1+c2c = {c_1} + {c_2} in the above equation
∴I=−9−x2 +3sin⁡−1x3+c\therefore I = - \sqrt {9 - {x^2}} \, + 3{\sin ^{ - 1}}\dfrac{x}{3} + c

Therefore, the integral of [x+39−x2]\left[ {\dfrac{{x + 3}}{{\sqrt {9 - {x^2}} }}} \right] is −9−x2 +3sin⁡−1x3+c - \sqrt {9 - {x^2}} \, + 3{\sin ^{ - 1}}\dfrac{x}{3} + c.

Note: We can also do this question by substituting x with trigonometric functions.Then we don’t have to split the terms. But this is the easiest method to answer this question.Here we have just applied a simple substitution and a standard formula.