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Question: How do you find integral of \(\dfrac{1}{{\sqrt {1 - 4{x^2}} }}dx?\)...

How do you find integral of 11−4x2dx?\dfrac{1}{{\sqrt {1 - 4{x^2}} }}dx?

Explanation

Solution

Here we have an indefinite integral. The term inside the integral sign is called the integrand. We can solve this using the integration formula ∫dxa2−x2=sin⁡−1xa+c\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\dfrac{x}{a} + c} . To apply this formula we need to simplify the integrand. Because we need a coefficient of ‘x’ to be 1. After that we can apply this integral formula.

Complete step-by-step solution:
Given,
∫11−4x2dx\int {\dfrac{1}{{\sqrt {1 - 4{x^2}} }}dx}
As we said earlier we need the coefficient of ‘x’ to be 1. So take 4 common,
=∫14(14−x2)dx= \int {\dfrac{1}{{\sqrt {4\left( {\dfrac{1}{4} - {x^2}} \right)} }}dx}
=∫1414−x2dx= \int {\dfrac{1}{{\sqrt 4 \sqrt {\dfrac{1}{4} - {x^2}} }}} dx
We know 4 is a perfect square,
=∫1214−x2dx= \int {\dfrac{1}{{2\sqrt {\dfrac{1}{4} - {x^2}} }}} dx
Using the constant coefficient rule, we have
=12∫114−x2dx= \dfrac{1}{2}\int {\dfrac{1}{{\sqrt {\dfrac{1}{4} - {x^2}} }}} dx
Now Apply the formula ∫dxa2−x2=sin⁡−1xa+c\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\dfrac{x}{a} + c}
=12[sin⁡−1x(12)+4]+c= \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\dfrac{x}{{\left( {\dfrac{1}{2}} \right)}} + 4} \right] + c
=12sin⁡−12x+c= \dfrac{1}{2}{\sin ^{ - 1}}2x + c
∫11−4x2dx=12sin⁡−12x+c\int {\dfrac{1}{{\sqrt {1 - 4{x^2}} }}dx} = \dfrac{1}{2}{\sin ^{ - 1}}2x + c
Where ‘c’ is integral constant.

Hence the final answer is ∫11−4x2dx=12sin⁡−12x+c\int {\dfrac{1}{{\sqrt {1 - 4{x^2}} }}dx} = \dfrac{1}{2}{\sin ^{ - 1}}2x + c

Note: In the given above problem we have an indefinite integral, that is no upper and lower limit. Hence we add the integration constant ‘c’ after integrating. In a definite integral we will have an upper and lower limit, we don’t need to add integration constant in the case of definite integral. We have different integration rule:
The power rule: If we have a variable ‘x’ raised to a power ‘n’ then the integration is given by ∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} .
The constant coefficient rule: if we have an indefinite integral of K.f(x)K.f(x), where f(x) is some function and ‘K’ represents a constant then the integration is equal to the indefinite integral of f(x) multiplied by ‘K’. That is ∫K.f(x)dx=K∫f(x)dx\int {K.f(x)dx = K\int {f(x)dx} } .
The sum rule: if we have to integrate functions that are the sum of several terms, then we need to integrate each term in the sum separately. That is
∫(f(x)+g(x))dx=∫f(x)dx+∫g(x)dx\int {\left( {f(x) + g(x)} \right)dx = \int {f(x)dx} } + \int {g(x)dx}
For the difference rule we have to integrate each term in the integrand separately.