Question: find the value of given integral. ( ( tanx)^4)×((tanx -1)^4) from lower limit =0 to upper limit=π÷4
I=∫_0^(π/4)tan^4x (tanx-1)^4 dx
Let u=tanx, then du/dx=sec^2x=1+tan^2x=1+u^2
i.e. dx=du/(1+u^2 )
Ignoring the limits, for the moment.
I=∫u^4 (u-1)^4∙du/(1+u^2 )
I=∫(u^4 (u-1)^4)/(1+u^2 )∙du
In the expression, (u^4 (u-1)^4)/(1+u^2 ), divide the numerator through by the denominator, (I.e. long division of polynomials) to get,
(u^4 (u-1)^4)/(1+u^2 )=u^6 - 4u^5 + 5u^4 - 4u^2 + 4 - 4/(1+u^2 )
Then,
I=∫u^6 - 4u^5 + 5u^4 - 4u^2 + 4 - 4/(1+u^2 ) du
I=∫u^6 - 4u^5 + 5u^4 - 4u^2 + 4 du - 4∫1/(1+u^2 ) du
I=1/7 u^7 - 2/3 u^6 + u^5 - 4/3 u^3 + 4u - 4∫1/(1+u^2 ) du
And, using the initial substitution,
4∫du/(1+u^2 )=4∫1 dx=4x
Hence,
I=1/7 u^7-2/3 u^6+u^5-4/3 u^3+4u-4x
I=1/7.tan^7(x) - 2/3.tan^6(x) + tan^5(x) - 4/3.tan^3(x) + 4tan(x) - 4x
Now inputting the limits,
I=[1/7.tan^7(x) - 2/3.tan^6(x) + tan^5(x) - 4/3.tan^3(x) + 4tan(x) - 4x]_0^(π/4)
I={(1/7 - 2/3 + 1 - 4/3 + 4 - 4.π/4) - (0-0+0-0+0-0) }
I = 22/7 - π≅0