# The value of the integral

Question:

The value of the integral $\int_{0}^{1} x \cot ^{-1}\left(1-x^{2}+x^{4}\right) d x$ is:

1. (1) $\frac{\pi}{2}-\frac{1}{2} \log _{e} 2$

2. (2) $\frac{\pi}{4}-\log _{e} 2$

3. (3) $\frac{\pi}{2}-\log _{e} 2$

4. (4) $\frac{\pi}{4}-\frac{1}{2} \log _{e} 2$

Correct Option: , 4

Solution:

$\int_{0}^{1} x \cot ^{-1}\left(1-x^{2}+x^{4}\right) d x=\int_{0}^{1} x \tan ^{-1}\left(\frac{1}{1+x^{4}-x^{2}}\right)$

$=\int_{0}^{1} x \tan ^{-1}\left(\frac{x^{2}-\left(x^{2}-1\right)}{1+x^{2}\left(x^{2}-1\right)}\right) d x$

$=\frac{1}{2} \int_{0}^{1} 1 \tan ^{-1} t^{2} d t-\frac{1}{2} \int_{-1}^{0} 1 \tan ^{-1} k d k$

Put $x^{2}=t \Rightarrow 2 x d x=d t$ in the first integral and $x^{2}-1=k \Rightarrow 2 x d x=d k$ in the second integral.

$=\frac{1}{2} \int_{0}^{1} 1 \tan ^{-1} t d t-\frac{1}{2} \int_{0}^{1} 1 \tan ^{-1} k d k$

$=\frac{1}{2}\left(\left.t \tan ^{-1} t\right|_{0} ^{1}-\int_{0}^{1} \frac{t}{1+t^{2}} d t\right)$

$-\frac{1}{2}\left(\left.k \tan ^{-1} k\right|_{0} ^{1}-\int_{-1}^{0} \frac{k}{1+k^{2}} d k\right)$

$=\frac{1}{2}\left(\frac{\pi}{4}-\left(\left.\frac{1}{2} \ln \left(1+t^{2}\right)\right|_{0} ^{1}\right)-\frac{1}{2}\left(-\frac{\pi}{4}-\left(\left.\frac{1}{2} \ln \left(1+k^{2}\right)\right|_{-1} ^{0}\right)\right)\right.$

$=\left(\frac{\pi}{8}-\frac{1}{4} \ln 2\right)-\left(\frac{-\pi}{8}-\frac{1}{4} 10-\ln 2\right)=\frac{\pi}{4}-\frac{1}{2} \ln 2$