Question

If \[\int {\dfrac{{\left( {\cos x -\ sin x} \right)}}{{\sqrt {8 - \sin 2x} }}} dx = a\sin^{ - 1}\left( {\dfrac{{\sin x +\cos x}}{b}} \right) + c\] where \[c\] is a constant of integration. Then what is the value of ordered pair \[\left( {a,b} \right)\]?
A. \[\left( {1, - 3} \right)\]
B. \[\left( {1,3} \right)\]
C. \[\left( { - 1,3} \right)\]
D. \[\left( {3,1} \right)\]

Answer 1

Hint: To solve the question we will assume that \[\sin x + \cos x = u\], then find it derivative. After that, we will take square both sides of the equation \[\sin x + \cos x = u\] and calculate the value of \[\sin 2x\] in term of \[u\]. Then substitute the value of \[\sin 2x\] and \[dx\] in the integration \[\int {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\sqrt {8 - \sin 2x} }}} dx\]. Then we will apply the formula \[\int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}d} x = si{n^{ - 1}}\left( {\dfrac{x}{a}} \right) + c\] and substitute the value of \[u\] and compare the result with \[asi{n^{ - 1}}\left( {\dfrac{{sinx + cosx}}{b}} \right) + c\].
Formula used :
Derivative formula:
\[\dfrac{d}{{dx}}\left( {sinx} \right) = cosx\]
\[\dfrac{d}{{dx}}\left( {cosx} \right) = - sinx\]
Integration formula
\[\int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}} dx = si{n^{ - 1}}\left( {\dfrac{x}{a}} \right) + c\]
Identity formula:
\[si{n^2}x + co{s^2}x = 1\]
Complete step by step solution:
The given integral equation is \[\int {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\sqrt {8 - \sin 2x} }}} dx = a\sin^{ - 1}\left( {\dfrac{{\sin x + \cos x}}{b}} \right) + c\].
Let’s solve the above integral.
Let consider,
\[I = \int {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\sqrt {8 - \sin 2x} }}} dx\] \[.....\left( 1 \right)\]
Apply the substitution method of integration.
Put \[\sin x + \cos x = u\] \[.....\left( 2 \right)\]
Differentiate the above equation with respect to \[x\].
\[\cos x - \sin x = \dfrac{{du}}{{dx}}\]
\[ \Rightarrow \]\[\left( {\cos x - \sin x} \right)dx = du\]

Take square of equation \[\left( 2 \right)\].
\[{\left( {\sin x + \cos x} \right)^2} = {u^2}\]
\[ \Rightarrow \]\[\sin^{2} x + \cos^{2} x + 2\sin x\cos x = {u^2}\] [Since \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]]
\[ \Rightarrow \]\[1 + \sin 2x = {u^2}\] [Since \[\sin^{2} x + \cos^{2} x = 1\] and \[2\sin x\cos x = \sin 2x\]]
\[ \Rightarrow \]\[\sin 2x = {u^2} - 1\] \[.....\left( 3 \right)\]

Now substitute equations \[\left( 2 \right)\] and \[\left( 3 \right)\] in the equation \[\left( 1 \right)\].
\[I = \int {\dfrac{1}{{\sqrt {8 - \left( {{u^2} - 1} \right)} }}} du\]
\[ \Rightarrow \]\[I = \int {\dfrac{1}{{\sqrt {8 - {u^2} + 1} }}} du\]
\[ \Rightarrow \]\[I = \int {\dfrac{1}{{\sqrt {9 - {u^2}} }}} du\]
\[ \Rightarrow \]\[I = \int {\dfrac{1}{{\sqrt {{3^2} - {u^2}} }}} du\]

Now apply the formula \[\int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}} dx = \sin^{ - 1}\left( {\dfrac{x}{a}} \right) + c\].
\[I = \sin^{ - 1}\left( {\dfrac{u}{3}} \right) + c\]
Resubstitute the value of \[u\].
\[I = \sin^{ - 1}\left( {\dfrac{{\sin x + \cos x}}{3}} \right) + c\]
Now compare the above equation with the right-hand side of the given integral equation.
Then,
\[a = 1, b = 3\]
Thus, the coordinates of the ordered pair \[\left( {a,b} \right)\] are \[\left( {1,3} \right)\].
Hence the correct option is B.
Note: The integration is the process of finding the antiderivative of the function. We solve the given integral by substitution method of integration. This method is also called as reverse chain rule or a U-substitution method.
Steps of substitution method:
Choose the new variable and make the substitution
Integrate the given function with respect to the substituted variable.
Resubstitute the original value of the variable.