Question

How do find $\int {\dfrac{1}{{\sqrt { - {x^2} - 4x} }}dx} ?$

Answer 1

Hint:To find the answer of given question, you have to integrate the given function and in order to integrate it, first make the expression under the square root a perfect square with help of algebraic operations and then use substitution method to simplify or make the function to be integrated easily (substitute any trigonometric function). And at last put the substitution back and have the required answer.

Complete step by step answer:
In order to find the integration of the given integral $\int {\dfrac{1}{{\sqrt { - {x^2} - 4x} }}dx} $, we will first simplify the expression under the root as a perfect square as following.
We can write the expression under the root as
$- {x^2} - 4x \\
\Rightarrow - {x^2} - 2 \times 2x \\ $
Adding and subtracting four in order to make it perfect square, we will get
$- {x^2} - 2 \times 2x + 4 - 4 \\
\Rightarrow - {x^2} - 2 \times 2x - 4 + 4 \\
\Rightarrow - ({x^2} + 2 \times 2x + 4) + 4 \\
\Rightarrow 4 - {(x + 2)^2} \\ $
Now coming to the integration, we can now write the integration as
$\int {\dfrac{1}{{\sqrt { - {x^2} - 4x} }}dx} = \int {\dfrac{1}{{\sqrt {4 - {{(x + 2)}^2}} }}dx} $
We will proceed to integrate after substitution method, we will substitute $(x + 2) = 2\sin \theta $
Differentiating both sides, we will get
$dx = 2\cos \theta d\theta $
Now, substituting these, we will get
\[\int {\dfrac{1}{{\sqrt {4 - {{(x + 2)}^2}} }}dx} = \int {\dfrac{{2\cos \theta }}{{\sqrt {4 - {{(2\sin \theta )}^2}} }}d\theta } \]
Solving this further,
\[\int {\dfrac{{2\cos \theta }}{{\sqrt {4 - {{(2\sin \theta )}^2}} }}d\theta } = \int {\dfrac{{2\cos \theta }}{{\sqrt {4 - 4{{\sin }^2}\theta } }}d\theta } \\
\Rightarrow\int {\dfrac{{2\cos \theta }}{{\sqrt {4 - {{(2\sin \theta )}^2}} }}d\theta }= \int {\dfrac{{2\cos \theta }}{{2\sqrt {1 - {{\sin }^2}\theta } }}d\theta } \\
\Rightarrow\int {\dfrac{{2\cos \theta }}{{\sqrt {4 - {{(2\sin \theta )}^2}} }}d\theta }= \int {\dfrac{{\cos \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}d\theta } \\ \]
Now from the trigonometric relations of sine and cosine, we know that \[\cos \theta = \sqrt {1 - {{\sin }^2}\theta } \]
Using this to solve further, we will get
\[\int {\dfrac{{\cos \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}d\theta } = \int {\dfrac{{\cos \theta }}{{\cos \theta }}d\theta } \\
\Rightarrow\int {\dfrac{{\cos \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}d\theta }= \int {d\theta } \\
\Rightarrow\int {\dfrac{{\cos \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}d\theta }= \theta + c \\ \]
So we have considered $(x + 2) = 2\sin \theta $,
$(x + 2) = 2\sin \theta \\
\Rightarrow \sin \theta = \dfrac{{(x + 2)}}{2} \\
\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{{x + 2}}{2}} \right) \\ $
Putting the value of $\theta $ in the integral, we will get
$\int {\dfrac{1}{{\sqrt { - {x^2} - 4x} }}dx} = \theta + c = {\sin ^{ - 1}}\left( {\dfrac{{x + 2}}{2}} \right) + c$

Therefore ${\sin ^{ - 1}}\left( {\dfrac{{x + 2}}{2}} \right) + c$ is the required integration of the given function.

Note:When solving indefinite integral, always write the constant part after the integration of any function, because in indefinite integral the constant part is important but in definite integral the constant part is being cancelled out in the integration when putting the limits.