Question

If \[\int\limits_{ - a}^a {\sqrt {\dfrac{{a - x}}{{a + x}}} dx = k\pi } \], then what is the value of \[k\]?
A. \[ - a\]
B. \[ - 2a\]
C. \[2a\]
D. \[a\]

Answer 1

Hint: Here, an equation of the definite integral is given. First, substitute \[x = a\cos \theta \] in the given integral. Then, change the values of the limits according to the substituted value. After that, simplify the integral by using the trigonometric identities. Then, solve the integrals and apply the limits to calculate the value of the integral. In the end, compare the value of the integral with the given equation of the integral and get the required answer.

Formula Used: \[\sin 2x = 2\sin x\cos x\]
\[{\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}\]
\[{\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\]
\[\int {\cos xdx = \sin x + c} \]
\[\int {ndx = nx + c} \]
Integration Rule: \[\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx\]

Complete step by step solution: The given equation of the integral is \[\int\limits_{ - a}^a {\sqrt {\dfrac{{a - x}}{{a + x}}} dx = k\pi } \].
Let consider,
\[I = \int\limits_{ - a}^a {\sqrt {\dfrac{{a - x}}{{a + x}}} dx} \]
Substitute \[x = a\cos \theta \] in the above integral.
Differentiate the substituted equation.
\[dx = a\left( { - \sin \theta } \right)d\theta \]
\[ \Rightarrow dx = - a\sin \theta d\theta \]
The values of the limits are changed as follows:
At \[ - a\] : \[ - a = a\cos \theta \] \[ \Rightarrow \cos \theta = - 1\] \[ \Rightarrow \theta = \pi \]
At \[a\] : \[a = a\cos \theta \] \[ \Rightarrow \cos \theta = 1\] \[ \Rightarrow \theta = 0\]

Now substitute all values in the above integral.
We get,
\[I = \int\limits_\pi ^0 {\sqrt {\dfrac{{a - a\cos \theta }}{{a + a\cos \theta }}} \left( { - a\sin \theta d\theta } \right)} \]
Simplify the integral.
\[I = - \int\limits_\pi ^0 {\sqrt {\dfrac{{a\left( {1 - \cos \theta } \right)}}{{a\left( {1 + \cos \theta } \right)}}} a\sin \theta d\theta } \]
Apply the integration rule \[\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx\] .
\[I = \int\limits_0^\pi {\sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} a\sin \theta d\theta } \]
Simplify the trigonometric terms by applying the trigonometric formulas \[\sin 2x = 2\sin x\cos x\], \[{\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}\] and \[{\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\].
We get,
\[I = \int\limits_0^\pi {a\sqrt {\dfrac{{2{{\sin }^2}\dfrac{\theta }{2}}}{{2{{\cos }^2}\dfrac{\theta }{2}}}} \left( {2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \right)d\theta } \]
Solve the square root.
\[ \Rightarrow I = 2a\int\limits_0^\pi {\left[ {\dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}\left( {\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \right)} \right]d\theta } \]
\[ \Rightarrow I = 2a\int\limits_0^\pi {{{\sin }^2}\dfrac{\theta }{2}d\theta } \]
Again, apply the trigonometric identity \[{\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\].
\[ \Rightarrow I = 2a\int\limits_0^\pi {\left[ {\dfrac{{1 - \cos \theta }}{2}} \right]d\theta } \]
\[ \Rightarrow I = a\int\limits_0^\pi {\left[ {1 - \cos \theta } \right]d\theta } \]
Now solve the integrals.
\[ \Rightarrow I = a\left[ {\theta - \sin \theta } \right]_0^\pi \]
Apply the upper and lower limits.
\[ \Rightarrow I = a\left[ {\left( {\pi - \sin \pi } \right) - \left( {0 - \sin 0} \right)} \right]\]
\[ \Rightarrow I = a\left[ {\left( {\pi - 0} \right) - \left( {0 - 0} \right)} \right]\]
\[ \Rightarrow I = a\pi \]
Thus, \[\int\limits_{ - a}^a {\sqrt {\dfrac{{a - x}}{{a + x}}} dx} = a\pi \].
Now compare the above equation with the originally given equation.
We get,
\[\int\limits_{ - a}^a {\sqrt {\dfrac{{a - x}}{{a + x}}} dx} = a\pi = k\pi \]
\[ \Rightarrow k = a\]

Option ‘D’ is correct

Note: Students often get confused and solve the integral as \[\int {\cos xdx = - \sin x + c} \]. They got confused because the derivative of \[\cos x\] is \[ - \sin x\]. But the correct formula is \[\int {\cos xdx = \sin x + c} \].