JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 1)
If $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c$$, where c is a constant of integration, then
the ordered pair (a, b) is equal to :
(-1, 3)
(1, 3)
(1, -3)
(3, 1)
Explanation
Given $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx$$
Write sin2x = 1 + sin2x - 1
= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx$$
= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \right]} }}} dx$$
= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} \right)}^2} - 1} \right]} }}} dx$$
= $$\int {{{\cos x - \sin x} \over {\sqrt {9 - {{\left( {\sin x + \cos x} \right)}^2}} }}} dx$$
put sin x + cos x = t
$$ \Rightarrow $$ (cos x – sin x) dx = dt
= $$\int {{{dt} \over {\sqrt {9 - {{\left( t \right)}^2}} }}} $$
= $${\sin ^{ - 1}}\left( {{t \over 3}} \right)$$ + C
= $${\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over 3}} \right) + C$$
$$ \therefore $$ a = 1 and b = 3
Write sin2x = 1 + sin2x - 1
= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx$$
= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \right]} }}} dx$$
= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} \right)}^2} - 1} \right]} }}} dx$$
= $$\int {{{\cos x - \sin x} \over {\sqrt {9 - {{\left( {\sin x + \cos x} \right)}^2}} }}} dx$$
put sin x + cos x = t
$$ \Rightarrow $$ (cos x – sin x) dx = dt
= $$\int {{{dt} \over {\sqrt {9 - {{\left( t \right)}^2}} }}} $$
= $${\sin ^{ - 1}}\left( {{t \over 3}} \right)$$ + C
= $${\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over 3}} \right) + C$$
$$ \therefore $$ a = 1 and b = 3
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