1
JEE Main 2019 (Online) 10th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let n $$ \ge $$ 2 be a natural number and $$0 < \theta < {\pi \over 2}.$$ Then $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$ is equal to - (where C is a constant of integration)
A
$${n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
B
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
C
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
D
$${n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
2
JEE Main 2019 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If   $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$

$$f\left( 0 \right) = 0,$$    then the value of $$f(1)$$ is :
A
$$ - $$ $${1 \over 2}$$
B
$$ - $$ $${1 \over 4}$$
C
$${1 \over 2}$$
D
$${1 \over 4}$$
3
JEE Main 2019 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For x2 $$ \ne $$ n$$\pi $$ + 1, n $$ \in $$ N (the set of natural numbers), the integral

$$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} $$ is equal to :

(where c is a constant of integration)
A
$${\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$$
B
$${1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c$$
C
$${1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c$$
D
$${\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + c$$
4
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}} $$ $$\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,$$ is a constant of integration) then the ordered pair (K, A) is equal to :
A
(2, 1)
B
($$-$$2, 3)
C
(2, 3)
D
($$-$$2, 1)
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