WB JEE 2010 | Indefinite Integrals Question 10 | Mathematics

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

$$\int {\sqrt {1 + \cos x} dx} $$ is equal to

$$2\sqrt 2 \cos {x \over 2} + C$$

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

$$\sqrt 2 \cos {x \over 2} + C$$

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

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

$$\int {{{{x^3}dx} \over {1 + {x^8}}} = } $$

$$4{\tan ^{ - 1}}{x^3} + c$$

$${1 \over 4}{\tan ^{ - 1}}{x^4} + c$$

$$x + 4{\tan ^{ - 1}}{x^4} + c$$

$${x^2} + {1 \over 4}{\tan ^{ - 1}}{x^4} + c$$

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

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

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

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

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

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

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