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Question

Question: $\int_{0}^{1}x \tan^{-1}x dx =$...

∫01xtan⁡−1xdx=\int_{0}^{1}x \tan^{-1}x dx =

A

π4+12\frac{\pi}{4}+\frac{1}{2}

B

π4−12\frac{\pi}{4}-\frac{1}{2}

C

12−π4\frac{1}{2}-\frac{\pi}{4}

D

−π4−12-\frac{\pi}{4}-\frac{1}{2}

Answer

π4−12\frac{\pi}{4}-\frac{1}{2}

Explanation

Solution

To evaluate the definite integral ∫01xtan⁡−1xdx\int_{0}^{1}x \tan^{-1}x dx, we will use the method of integration by parts.

The formula for integration by parts is ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

We choose u=tan⁡−1xu = \tan^{-1}x and dv=x dxdv = x \, dx. From these choices, we find dudu and vv:

du=ddx(tan⁡−1x) dx=11+x2 dxdu = \frac{d}{dx}(\tan^{-1}x) \, dx = \frac{1}{1+x^2} \, dx

v=∫x dx=x22v = \int x \, dx = \frac{x^2}{2}

Now, apply the integration by parts formula to the definite integral:

∫01xtan⁡−1xdx=[(tan⁡−1x)(x22)]01−∫01(x22)(11+x2) dx\int_{0}^{1}x \tan^{-1}x dx = \left[ (\tan^{-1}x) \left(\frac{x^2}{2}\right) \right]_{0}^{1} - \int_{0}^{1} \left(\frac{x^2}{2}\right) \left(\frac{1}{1+x^2}\right) \, dx

Let's evaluate the first term:

[x22tan⁡−1x]01=(122tan⁡−1(1))−(022tan⁡−1(0))\left[ \frac{x^2}{2} \tan^{-1}x \right]_{0}^{1} = \left( \frac{1^2}{2} \tan^{-1}(1) \right) - \left( \frac{0^2}{2} \tan^{-1}(0) \right)

Since tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4} and tan⁡−1(0)=0\tan^{-1}(0) = 0:

=(12⋅π4)−(0)=π8= \left( \frac{1}{2} \cdot \frac{\pi}{4} \right) - (0) = \frac{\pi}{8}

Now, let's evaluate the second term:

−∫01x22(1+x2) dx=−12∫01x21+x2 dx- \int_{0}^{1} \frac{x^2}{2(1+x^2)} \, dx = -\frac{1}{2} \int_{0}^{1} \frac{x^2}{1+x^2} \, dx

To simplify the integrand x21+x2\frac{x^2}{1+x^2}, we can rewrite it as:

x21+x2=1+x2−11+x2=1+x21+x2−11+x2=1−11+x2\frac{x^2}{1+x^2} = \frac{1+x^2-1}{1+x^2} = \frac{1+x^2}{1+x^2} - \frac{1}{1+x^2} = 1 - \frac{1}{1+x^2}

So the integral becomes:

−12∫01(1−11+x2) dx-\frac{1}{2} \int_{0}^{1} \left(1 - \frac{1}{1+x^2}\right) \, dx

=−12[x−tan⁡−1x]01= -\frac{1}{2} \left[ x - \tan^{-1}x \right]_{0}^{1}

Now, evaluate this expression at the limits:

=−12[(1−tan⁡−1(1))−(0−tan⁡−1(0))]= -\frac{1}{2} \left[ (1 - \tan^{-1}(1)) - (0 - \tan^{-1}(0)) \right]

=−12[(1−π4)−(0−0)]= -\frac{1}{2} \left[ (1 - \frac{\pi}{4}) - (0 - 0) \right]

=−12(1−π4)= -\frac{1}{2} \left(1 - \frac{\pi}{4}\right)

=−12+π8= -\frac{1}{2} + \frac{\pi}{8}

Finally, add the results of the first and second terms:

∫01xtan⁡−1xdx=π8+(−12+π8)\int_{0}^{1}x \tan^{-1}x dx = \frac{\pi}{8} + \left(-\frac{1}{2} + \frac{\pi}{8}\right)

=π8−12+π8= \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8}

=2π8−12= \frac{2\pi}{8} - \frac{1}{2}

=π4−12= \frac{\pi}{4} - \frac{1}{2}