Trigonometry
Inverse Trigonometric Functions
MCQ (Single Correct Answer)Subjective
1
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

$$\int {{{\cos 2x} \over {\cos x}}dx = } $$

A
$$2\sin x + \log |\sec x + \tan x| + C$$
B
$$2\sin x - \log |\sec x - \tan x| + C$$
C
$$2\sin x - \log |\sec x + \tan x| + C$$
D
$$2\sin x + \log |\sec x - \tan x| + C$$
2
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

$$\int {{{{{\sin }^8}x - {{\cos }^8}x} \over {1 - 2{{\sin }^2}x{{\cos }^2}x}}dx} $$

A
$$ - {1 \over 2}\sin 2x + C$$
B
$${1 \over 2}\sin 2x + C$$
C
$${1 \over 2}\sin x + C$$
D
$$ - {1 \over 2}\sin x + C$$
3
WB JEE 2011
MCQ (Single Correct Answer)
+1
-0.25

$$\int {{2^x}(f'(x) + f(x)\log 2)dx} $$ is

A
$${2^x}f'(x) + C$$
B
$${2^x}f(x) + C$$
C
$${2^x}(\log 2)f(x) + C$$
D
$$(\log 2)f(x) + C$$
4
WB JEE 2024
MCQ (Single Correct Answer)
+1
-0.25

$$ \text { If } \int \frac{\log _e\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\mathrm{f}(\mathrm{g}(x))+\mathrm{c} \text { then } $$

A
$$\mathrm{f}(x)=\frac{x^2}{2}, \mathrm{~g}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right)$$
B
$$\mathrm{f}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right), \mathrm{g}(x)=\frac{x^2}{2}$$
C
$$\mathrm{f}(x)=x^2, \mathrm{~g}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right)$$
D
$$\mathrm{f}(x)=\log _{\mathrm{e}}\left(x-\sqrt{1+x^2}\right), \mathrm{g}(x)=x^2$$
WB JEE Subjects