JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 16)
If $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C$$
(where C is a constant of integration), then the ordered pair (A, B) is equal to :
(where C is a constant of integration), then the ordered pair (A, B) is equal to :
(2, 1)
($$-$$ 2, $$-$$1)
($$-$$ 2, 1)
(2, $$-$$1)
Explanation
We can write,
7 - 6x - x2 = 16 - (x + 3)2
and $${d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)$$
So, $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
= $$\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
= $$\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$ -
$$\int {{1 \over {\sqrt {16 - {{(x + 3)}^2}} }}} dx$$
= $$\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$ -
$$\int {{1 \over {\sqrt {{{\left( 4 \right)}^2} - {{(x + 3)}^2}} }}} dx$$
= $$ - 2\sqrt {7 - 6x - {x^2}} $$ - $${\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right)$$ + C
By comparing with the given equation in the question we get,
A = - 2 and B = - 1
7 - 6x - x2 = 16 - (x + 3)2
and $${d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)$$
So, $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
= $$\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
= $$\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$ -
$$\int {{1 \over {\sqrt {16 - {{(x + 3)}^2}} }}} dx$$
= $$\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$ -
$$\int {{1 \over {\sqrt {{{\left( 4 \right)}^2} - {{(x + 3)}^2}} }}} dx$$
= $$ - 2\sqrt {7 - 6x - {x^2}} $$ - $${\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right)$$ + C
By comparing with the given equation in the question we get,
A = - 2 and B = - 1
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