Mathematics - Definite Integration Question with Solution | TestHub

Let:

$$I={\int }_0^{\frac{\pi }{4}}4x\left(\frac{\pi }{4}-x\right)\ln (1+\tan x)dx$$

Using the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$, we can rewrite the integral as:

$$I={\int }_0^{\frac{\pi }{4}}4\left(\frac{\pi }{4}-x\right)x\ln \left(1+\tan \left(\frac{\pi }{4}-x\right)\right)dx$$

We know that $\tan \left(\frac{\pi }{4}-x\right)=\frac{1-\tan x}{1+\tan x}$.

Therefore, $1+\tan \left(\frac{\pi }{4}-x\right)=1+\frac{1-\tan x}{1+\tan x}=\frac{1+\tan x+1-\tan x}{1+\tan x}=\frac{2}{1+\tan x}$.

Substituting this back into the integral:

$$I={\int }_0^{\frac{\pi }{4}}4x\left(\frac{\pi }{4}-x\right)\ln \left(\frac{2}{1+\tan x}\right)dx$$

Using the logarithm property $\ln \left(\frac{a}{b}\right)=\ln a-\ln b$:

$$I={\int }_0^{\frac{\pi }{4}}4x\left(\frac{\pi }{4}-x\right)(\ln 2-\ln (1+\tan x))dx$$

$$I={\int }_0^{\frac{\pi }{4}}4x\left(\frac{\pi }{4}-x\right)\ln 2dx-{\int }_0^{\frac{\pi }{4}}4x\left(\frac{\pi }{4}-x\right)\ln (1+\tan x)dx$$

Notice that the second term on the right side is the original integral $I$.

So, we have:

$$I=4\ln 2{\int }_0^{\frac{\pi }{4}}x\left(\frac{\pi }{4}-x\right)dx-I$$

Now, we can solve for $I$:

$$2I=4\ln 2{\int }_0^{\frac{\pi }{4}}\left(\frac{\pi }{4}x-x^2\right)dx$$

$$I=2\ln 2{\int }_0^{\frac{\pi }{4}}\left(\frac{\pi }{4}x-x^2\right)dx$$

Evaluate the integral:

$${\int }_0^{\frac{\pi }{4}}\left(\frac{\pi }{4}x-x^2\right)dx={\left[\frac{\pi }{4}\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^{\frac{\pi }{4}}$$

$$=\left(\frac{\pi }{4}\frac{(\frac{\pi }{4}{)}^2}{2}-\frac{(\frac{\pi }{4}{)}^3}{3}\right)-(0)$$

$$=\frac{\pi }{4}\frac{{\pi }^2}{32}-\frac{{\pi }^3}{192}$$

$$=\frac{{\pi }^3}{128}-\frac{{\pi }^3}{192}$$

To combine these terms, find a common denominator, which is 384:

$$=\frac{3{\pi }^3}{384}-\frac{2{\pi }^3}{384}=\frac{{\pi }^3}{384}$$

Substitute this back into the expression for $I$:

$$I=2\ln 2\left(\frac{{\pi }^3}{384}\right)$$

$$I=\frac{{\pi }^3}{192}\ln 2$$