Solveeit Logo
Login

Question

Question: How do you find the integral of \[\dfrac{1}{{{x^2}\sqrt {25 - {x^2}} }}dx\] ?...

How do you find the integral of 1x225−x2dx\dfrac{1}{{{x^2}\sqrt {25 - {x^2}} }}dx ?

Explanation

Solution

Hint : We need to evaluate ∫1x225−x2dx\int {\dfrac{1}{{{x^2}\sqrt {25 - {x^2}} }}dx} . We know that the term inside the integral sign is called integrand. We simplify the integrand by making a substitution that is we put x=5sin⁡tx = 5\sin t . After simplification we apply the integral. In the final we need the solution in terms of ‘x’ only. Here we have an indefinite integral.

Complete step-by-step answer :
Given,
∫1x225−x2dx\int {\dfrac{1}{{{x^2}\sqrt {25 - {x^2}} }}dx}
to simplify the integrand let’s put x=5sin⁡tx = 5\sin t
Differentiating with respect to ‘x’ we have,
dx=5cos⁡t.dtdx = 5\cos t.dt
Substituting these in the integral we have,
∫1x225−x2dx=∫1(5sin⁡t)225−(5sin⁡t)25cos⁡t.dt\int {\dfrac{1}{{{x^2}\sqrt {25 - {x^2}} }}dx} = \int {\dfrac{1}{{{{(5\sin t)}^2}\sqrt {25 - {{(5\sin t)}^2}} }}5\cos t.dt}
=∫1(5sin⁡t)225−(5sin⁡t)25cos⁡t.dt= \int {\dfrac{1}{{{{(5\sin t)}^2}\sqrt {25 - {{(5\sin t)}^2}} }}5\cos t.dt}
=∫125sin⁡2t25−25sin⁡2t5cos⁡t.dt= \int {\dfrac{1}{{25{{\sin }^2}t\sqrt {25 - 25{{\sin }^2}t} }}5\cos t.dt}
=∫125sin⁡2t25(1−sin⁡2t)5cos⁡t.dt= \int {\dfrac{1}{{25{{\sin }^2}t\sqrt {25(1 - {{\sin }^2}t)} }}5\cos t.dt}
We know 25 is a perfect square and we take it outside,
=∫5cos⁡t25×5sin⁡2t(1−sin⁡2t)dt= \int {\dfrac{{5\cos t}}{{25 \times 5{{\sin }^2}t\sqrt {(1 - {{\sin }^2}t)} }}dt}
We know from Pythagoras trigonometric identity that is sin⁡2t+cos⁡2t=1{\sin ^2}t + {\cos ^2}t = 1 , we have cos⁡t=1−sin⁡2t\cos t = \sqrt {1 - {{\sin }^2}t}
=∫5cos⁡t25×5sin⁡2t.cos⁡tdt= \int {\dfrac{{5\cos t}}{{25 \times 5{{\sin }^2}t.\cos t}}dt}
Cancelling terms we have,
=∫125sin⁡2tdt= \int {\dfrac{1}{{25{{\sin }^2}t}}dt}
=125∫1sin⁡2tdt= \dfrac{1}{{25}}\int {\dfrac{1}{{{{\sin }^2}t}}dt}
We know that ∫1sin⁡2tdt=−cot⁡x+c\int {\dfrac{1}{{{{\sin }^2}t}}dt} = - \cot x + c .
=−125cot⁡t+c= - \dfrac{1}{{25}}\cot t + c , where ‘c’ is the integration constant.
But we need the answers in terms of ‘x’ only. We know that cotangent is the ratio of cosine to sine.
=−125cos⁡tsin⁡t+c −−−−(1)= - \dfrac{1}{{25}}\dfrac{{\cos t}}{{\sin t}} + c{\text{ }} - - - - (1)
We have x=5sin⁡tx = 5\sin t
sin⁡t=x5 −−−(2)\sin t = \dfrac{x}{5}{\text{ }} - - - (2)
Now to find cos⁡t\cos t we have cos⁡t=1−sin⁡2t\cos t = \sqrt {1 - {{\sin }^2}t}
cos⁡t=1−(x5)2\cos t = \sqrt {1 - {{\left( {\dfrac{x}{5}} \right)}^2}}
cos⁡t=1−x225\cos t = \sqrt {1 - \dfrac{{{x^2}}}{{25}}}
cos⁡t=25−x225\cos t = \sqrt {\dfrac{{25 - {x^2}}}{{25}}}
cos⁡t=25−x25 −−−−(3)\cos t = \dfrac{{\sqrt {25 - {x^2}} }}{5}{\text{ }} - - - - (3)
Substituting equation (2) and (3) in equation (1) we have,
=−125(25−x25)(x5)+c= - \dfrac{1}{{25}}\dfrac{{\left( {\dfrac{{\sqrt {25 - {x^2}} }}{5}} \right)}}{{\left( {\dfrac{x}{5}} \right)}} + c
=−12525−x2x+c= - \dfrac{1}{{25}}\dfrac{{\sqrt {25 - {x^2}} }}{x} + c
Thus we have,
∫1x225−x2dx=−12525−x2x+c\int {\dfrac{1}{{{x^2}\sqrt {25 - {x^2}} }}dx} = - \dfrac{1}{{25}}\dfrac{{\sqrt {25 - {x^2}} }}{x} + c , where ‘c’ is the integration constant.
So, the correct answer is “ −12525−x2x+c - \dfrac{1}{{25}}\dfrac{{\sqrt {25 - {x^2}} }}{x} + c”.

Note : Here we have an indefinite integral that is no upper limit and lower limit. Hence, in the case of indefinite integral we have integration constant. In definite integral we have lower limits and upper limits. Hence, in the case of definite integral we don’t have integration constant. As we can see in the above problem by using the substitute rule we can simplify the problem easily.