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JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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\int\limits_{{-1}}^1 \sqrt{\cos x} \log _e\left(\frac{1-x}{1+x}\right) d x=

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Correct option: A

0

Option A, 0, is correct because the integrand is an odd function, and the integral is evaluated over a symmetric interval [-1, 1].

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\int \nolimits_0^\infty \,\,\log \,\,\left( {x + \frac{1}{x}} \right)\,\,\frac{{dx}}{{1 + {x^2}}}

f\left( x \right)

is differentiable and

\int \nolimits_0^{{t^2}} \,\,x\;\,f\left( x \right)\,\,dx = \frac{2}{5}\,\,{t^5}

f\left( {\frac{4}{{25}}} \right)

equals The value of

\int\limits_{{0}}^{11}[x]^{3} . \mathrm{dx}

[*]

denotes the greatest integer function, is The integral

\int\limits_{2}^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x

is equal to: The definite integral

\int\limits_{{-2}}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x

I\,\,\left( {m,\,\,n} \right) = \int \nolimits_0^1 \,\,{t^m}\,\,{\left( {1 + t} \right)^n}\,\,dt

,then the expression for

I\left( {m,\,\,n} \right)

I\,\,\left( {m + 1,\,\,n + 1} \right)