1
JEE Main 2021 (Online) 18th March Morning Shift
+4
-1
The integral $$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$$ is equal to (where c is a constant of integration)
A
$${1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
B
$${1 \over 2}\cos \sqrt {{{(2x + 1)}^2} + 5} + c$$
C
$${1 \over 2}\cos \sqrt {{{(2x - 1)}^2} + 5} + c$$
D
$${1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c$$
2
JEE Main 2021 (Online) 25th February Evening Shift
+4
-1
The integral $$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx$$, x > 0, is equal to : (where c is a constant of integration)
A
$${\log _e}\sqrt {{x^2} + 5x - 7} + c$$
B
$$4{\log _e}|{x^2} + 5x - 7| + c$$
C
$${1 \over 4}{\log _e}|{x^2} + 5x - 7| + c$$
D
$${\log _e}|{x^2} + 5x - 7| + c$$
3
JEE Main 2021 (Online) 25th February Morning Shift
+4
-1
The value of the integral
$$\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta$$ is :
A
$${1 \over {18}}{\left[ {9 - 2{{\cos }^6}\theta - 3{{\cos }^4}\theta - 6{{\cos }^2}\theta } \right]^{{3 \over 2}}} + c$$
B
$${1 \over {18}}{\left[ {11 - 18{{\sin }^2}\theta + 9{{\sin }^4}\theta - 2{{\sin }^6}\theta } \right]^{{3 \over 2}}} + c$$
C
$${1 \over {18}}{\left[ {11 - 18{{\cos }^2}\theta + 9{{\cos }^4}\theta - 2{{\cos }^6}\theta } \right]^{{3 \over 2}}} + c$$
D
$${1 \over {18}}{\left[ {9 - 2{{\sin }^6}\theta - 3{{\sin }^4}\theta - 6{{\sin }^2}\theta } \right]^{{3 \over 2}}} + c$$
4
JEE Main 2021 (Online) 24th February Morning Shift
+4
-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 :
A
(-1, 3)
B
(1, 3)
C
(1, -3)
D
(3, 1)
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