WB JEE 2011 | Indefinite Integrals Question 12 | Mathematics

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

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

$$2\sin x + \log |\sec x + \tan x| + C$$

$$2\sin x - \log |\sec x - \tan x| + C$$

$$2\sin x - \log |\sec x + \tan x| + C$$

$$2\sin x + \log |\sec x - \tan x| + C$$

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

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

$$ - {1 \over 2}\sin 2x + C$$

$${1 \over 2}\sin 2x + C$$

$${1 \over 2}\sin x + C$$

$$ - {1 \over 2}\sin x + C$$

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

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

$${2^x}f'(x) + C$$

$${2^x}f(x) + C$$

$${2^x}(\log 2)f(x) + C$$

$$(\log 2)f(x) + C$$

WB JEE 2024

MCQ (Single Correct Answer)

-0.25

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

$$\mathrm{f}(x)=\frac{x^2}{2}, \mathrm{~g}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right)$$

$$\mathrm{f}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right), \mathrm{g}(x)=\frac{x^2}{2}$$

$$\mathrm{f}(x)=x^2, \mathrm{~g}(x)=\log _{\mathrm{e}}\left(x+\sqrt{1+x^2}\right)$$

$$\mathrm{f}(x)=\log _{\mathrm{e}}\left(x-\sqrt{1+x^2}\right), \mathrm{g}(x)=x^2$$

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