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Question: Evaluate the following integrals: \(\int {\sqrt {2 - 2x - 2{x^2}} } dx\)...

Evaluate the following integrals:
∫2−2x−2x2dx\int {\sqrt {2 - 2x - 2{x^2}} } dx

Explanation

Solution

First we will try to make the expression that has to be integrated into any well-known form, by using some formulas like a2+2ab+b2=(a+b)2{a^2} + 2ab + {b^2} = {(a + b)^2}, we will be able to make the expression in the form of a general expression whose integration formula is well defined i.e. of the form∫a2−x2dx\int {\sqrt {{a^2} - {x^2}} dx} and this integration is well defined as

∫a2−x2dx=xa2−x22+a22sin⁡−1(xa)+c\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{{x\sqrt {{a^2} - {x^2}} }}{2} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c

And now using this formula we will get our answer.

Complete step by step solution: Given data: ∫2−2x−2x2dx\int {\sqrt {2 - 2x - 2{x^2}} } dx

Simplifying the given term

⇒∫2−2x−2x2dx \Rightarrow \int {\sqrt {2 - 2x - 2{x^2}} } dx

Adding and subtracting12\dfrac{1}{2} in the root term

=∫2+12−12−2x−2x2dx = \int {\sqrt {2 + \dfrac{1}{2} - \dfrac{1}{2} - 2x - 2{x^2}} } dx

Now taking -1 common from the negative terms

=∫2+12−(12+2x+2x2)dx = \int {\sqrt {2 + \dfrac{1}{2} - \left( {\dfrac{1}{2} + 2x + 2{x^2}} \right)} } dx

By simplifying the terms of the brackets into the form of a2+2ab+b2{a^2} + 2ab + {b^2}

=∫2+12−((12)2+2(12)(2x)x+(2x)2)dx = \int {\sqrt {2 + \dfrac{1}{2} - \left( {{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} + 2\left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\sqrt 2 x} \right)x + {{\left( {\sqrt 2 x} \right)}^2}} \right)} } dx

Now, using a2+2ab+b2=(a+b)2{a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}

=∫2+12−((12)+(2x))2dx = \int {\sqrt {2 + \dfrac{1}{2} - {{\left( {\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\sqrt 2 x} \right)} \right)}^2}} } dx

Now let (12)+(2x)=t\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\sqrt 2 x} \right) = t

On differentiating with-respect-to x

⇒2dx=dt \Rightarrow \sqrt 2 dx = dt

⇒dx=dt2 \Rightarrow dx = \dfrac{{dt}}{{\sqrt 2 }}

Now substituting the value of (12)+(2x)\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\sqrt 2 x} \right)and dxdx

=∫2+12−t2dt2 = \int {\sqrt {2 + \dfrac{1}{2} - {t^2}} } \dfrac{{dt}}{{\sqrt 2 }}

Now let 2+12=a22 + \dfrac{1}{2} = {a^2}

=∫12a2−t2dt = \int {\dfrac{1}{{\sqrt 2 }}\sqrt {{a^2} - {t^2}} } dt

Now using ∫a2−x2dx=xa2−x22+a22sin⁡−1(xa)+c\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{{x\sqrt {{a^2} - {x^2}} }}{2} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c

⇒∫12a2−t2dt=12(ta2−t22+a22sin⁡−1(ta)+c) \Rightarrow \int {\dfrac{1}{{\sqrt 2 }}\sqrt {{a^2} - {t^2}} } dt = \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{t\sqrt {{a^2} - {t^2}} }}{2} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\dfrac{t}{a}} \right) + c} \right)

Now, simplifying the brackets

=ta2−t222+a222sin⁡−1(ta)+c2 = \dfrac{{t\sqrt {{a^2} - {t^2}} }}{{2\sqrt 2 }} + \dfrac{{{a^2}}}{{2\sqrt 2 }}{\sin ^{ - 1}}\left( {\dfrac{t}{a}} \right) + \dfrac{c}{{\sqrt 2 }}

Now substituting t=(12)+(2x)t = \left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\sqrt 2 x} \right)

=[12+2x]a2−[12+2x]222+a222sin⁡−1(12+2xa)+c2 = \dfrac{{\left[ {\dfrac{1}{{\sqrt 2 }} + \sqrt 2 x} \right]\sqrt {{a^2} - {{\left[ {\dfrac{1}{{\sqrt 2 }} + \sqrt 2 x} \right]}^2}} }}{{2\sqrt 2 }} + \dfrac{{{a^2}}}{{2\sqrt 2 }}{\sin ^{ - 1}}\left( {\dfrac{{\dfrac{1}{{\sqrt 2 }} + \sqrt 2 x}}{a}} \right) + \dfrac{c}{{\sqrt 2 }}

Now, simplifying the brackets

=(1+2x)a2−(12+2x2+2x)4+a222sin⁡−1(1+2x2a)+c2 = \dfrac{{\left( {1 + 2x} \right)\sqrt {{a^2} - \left( {\dfrac{1}{2} + 2{x^2} + 2x} \right)} }}{4} + \dfrac{{{a^2}}}{{2\sqrt 2 }}{\sin ^{ - 1}}\left( {\dfrac{{1 + 2x}}{{\sqrt 2 a}}} \right) + \dfrac{c}{{\sqrt 2 }}

Now substituting a2=2+12{a^2} = 2 + \dfrac{1}{2}or a=52a = \sqrt {\dfrac{5}{2}}

=(1+2x)2+12−(12+2x2+2x)4+542sin⁡−1(1+2x5)+c2 = \dfrac{{\left( {1 + 2x} \right)\sqrt {2 + \dfrac{1}{2} - \left( {\dfrac{1}{2} + 2{x^2} + 2x} \right)} }}{4} + \dfrac{5}{{4\sqrt 2 }}{\sin ^{ - 1}}\left( {\dfrac{{1 + 2x}}{{\sqrt 5 }}} \right) + \dfrac{c}{{\sqrt 2 }}

on simplifying the brackets

=(1+2x)2−2x2−2x4+542sin⁡−1(1+2x5)+c2 = \dfrac{{\left( {1 + 2x} \right)\sqrt {2 - 2{x^2} - 2x} }}{4} + \dfrac{5}{{4\sqrt 2 }}{\sin ^{ - 1}}\left( {\dfrac{{1 + 2x}}{{\sqrt 5 }}} \right) + \dfrac{c}{{\sqrt 2 }}

Now let c2=C\dfrac{c}{{\sqrt 2 }} = C

Therefore the required answer is (1+2x)2−2x2−2x4+542sin⁡−1(1+2x5)+C\dfrac{{\left( {1 + 2x} \right)\sqrt {2 - 2{x^2} - 2x} }}{4} + \dfrac{5}{{4\sqrt 2 }}{\sin ^{ - 1}}\left( {\dfrac{{1 + 2x}}{{\sqrt 5 }}} \right) + C

Note: Here we assumed the given variable is a new variable we get our answer in a new variable also, so some students just leave it in the form of a new variable only that is wrong as it is defined by us and not predefined so we always must resubstitute into the given variable as we should always be in the form of the terms given in the question.