If - 0 1 tan -1 x x dx-k- 0 - -2 x sin x dx then the value of k is

The correct option is C 1 Put x=tan a ∫_0^π/4sec^2aa/tan ada; put t=2a∫_0^π/2t/sin tdt k=1 ...

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Integration Using Substitution

If ∫ 0 1 ta...

Question

If $\int_{0}^{1} \frac{{tan}^{- 1} x}{x} d x = k \int_{0}^{π / 2} \frac{x}{sin x} d x$ then the value of $k$ is

1

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\frac{1}{4}

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4

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\frac{1}{2}

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Similar questions

I_{1} = \int_{0}^{1} \frac{{tan}^{- 1} x d x}{x}

I_{2} = \int_{0}^{\frac{π}{2}} \frac{x d x}{sin x}

\frac{I_{1}}{I_{2}}

\int_{0}^{1} \frac{{tan}^{- 1} x}{x} d x = k \int_{0}^{π / 2} \frac{x}{sin x} d x

k

\int_{0}^{π} x f ({sin}^{2} x + {sec}^{2} x) d x = k \int_{0}^{π / 2} f ({sin}^{2} x + {sec}^{2} x) d x

k

I = \int_{0}^{1} {cot}^{- 1} (1 - x + x^{2}) dx = k \int_{0}^{1} {tan}^{- 1} x dx

I = \int_{0}^{\frac{π}{4}} ({tan}^{n + 1} x) d x + \frac{1}{2} \int_{0}^{\frac{π}{2}} ({tan}^{n + 1} (\frac{x}{2})) d x

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