1
JEE Main 2023 (Online) 10th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

If $$I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx}$$ and $$I(0) = 1$$, then $$I\left( {{\pi \over 3}} \right)$$ is equal to :

A
$$- {e^{{3 \over 4}}}$$
B
$$- {1 \over 2}{e^{{3 \over 4}}}$$
C
$${e^{{3 \over 4}}}$$
D
$${1 \over 2}{e^{{3 \over 4}}}$$
2
JEE Main 2023 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

The integral $$\int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x$$ is equal to :

A
$$\left(\frac{x}{2}\right)^{x}+\left(\frac{2}{x}\right)^{x}+C$$
B
$$\left(\frac{x}{2}\right)^{x}-\left(\frac{2}{x}\right)^{x}+C$$
C
$$\left(\frac{x}{2}\right)^{x} \log _{2}\left(\frac{2}{x}\right)+C$$
D
None
3
JEE Main 2023 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let $$I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0$$. If $$\lim_\limits{x \rightarrow \infty} I(x)=0$$, then $$I(1)$$ is equal to :

A
$$\frac{e+1}{e+2}-\log _{e}(e+1)$$
B
$$\frac{e+1}{e+2}+\log _{e}(e+1)$$
C
$$\frac{e+2}{e+1}-\log _{e}(e+1)$$
D
$$\frac{e+2}{e+1}+\log _{e}(e+1)$$
4
JEE Main 2023 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let $$I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\left(\frac{\pi}{4}\right)$$ is equal to :

A
$$\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}$$
B
$$\log _{e} \frac{(\pi+4)^{2}}{16}-\frac{\pi^{2}}{4(\pi+4)}$$
C
$$\log _{e} \frac{(\pi+4)^{2}}{16}+\frac{\pi^{2}}{4(\pi+4)}$$
D
$$\log _{e} \frac{(\pi+4)^{2}}{32}+\frac{\pi^{2}}{4(\pi+4)}$$
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