1
JEE Main 2021 (Online) 18th March Morning Shift
MCQ (Single Correct Answer) +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)
2
JEE Main 2021 (Online) 25th February Evening Shift
MCQ (Single Correct Answer) +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)
3
JEE Main 2021 (Online) 25th February Morning Slot
MCQ (Single Correct Answer) +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 :
$$\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta $$ is :
4
JEE Main 2021 (Online) 24th February Morning Slot
MCQ (Single Correct Answer) +4 -1
English
Hindi
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 :
Questions Asked from Indefinite Integrals (MCQ (Single Correct Answer))
Number in Brackets after Paper Indicates No. of Questions
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