1
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin x}\,dx =$
A
$\log(\cos x) + \cos x + c$, where c is the constant of integration
B
$\log(\sin x) - \sin x + c$, where c is the constant of integration
C
$\log(\sin x) + \sin x + c$, where c is the constant of integration
D
$\log(\cos x) - \cos x + c$, where c is the constant of integration
2
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx =$
A
$0$
B
$\dfrac{\pi}{12}$
C
$\dfrac{\pi}{24}$
D
$\dfrac{\pi^2}{2}$
3
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int_0^{\log 5} \dfrac{e^x\sqrt{e^x - 1}}{e^x + 3}\,dx =$
A
$2 - \dfrac{\pi}{4}$
B
$4 - \dfrac{\pi}{4}$
C
$2 - \pi$
D
$4 - \pi$
4
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The integrating factor of the differential equation $(1 + t^2) + \left(x - e^{\tan^{-1}t}\right)\dfrac{dt}{dx} = 0$ is
A
$e^{\tan^{-1}t}$
B
$-e^{\tan^{-1}t}$
C
$e^{-\tan^{-1}t}$
D
$-e^{-\tan^{-1}t}$

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