Question

Evaluate integral from 0 to pi/4 2tan^3x dx

Answer 1

hope it helps u
concept:definite integration
sub tobic:trigonometry,differentiation ,basic integration
method used:substituiton method

Answer 2

The Answer is 1- log2.

1) $\int\limits^\frac{\pi}{4} _0 {2 tan^3 x} \, dx$

2) $2\int\limits^\frac{\pi}{4} _b {tan^2 x * tan x} \, dx$

3) $2\int\limits^\frac{\pi}{4} _0 {(sec^2 x -1)* tan x} \, dx$

4) Put  sec x = u

5) du/dx = sec x * tan x, dx = du/(sec x* tan x)

6) Substituting the values :   $2\int\limits^\frac{\pi}{4} _0 {(u^2 - 1)* tan x} \, du/(u * tan x)$

7) SImplifying $2\int\limits^\frac{\pi}{4} _0 {(u^2-1)/u} \, du$

8) $2\int\limits^\frac{\pi}{4} _0 {u - \frac{1}{u} } \, du$

9) $2\int\limits^\frac{\pi}{4} _0 {u } \, du - 2\int\limits^\frac{\pi}{4} _0 {\frac{1}{u} } \, du$

10) $2\left[\begin{array}{ccc}\frac{u^2}{2} \end{array}\right]$  - $2\left[\begin{array}{ccc}\log u \end{array}\right]$  with limits pi/ 4 to 0

11) $u^{2} - 2 log u$ with limits pi/4 to 0

12) Substituting back : $sec^2 x - 2 log (sec x)$ with limits pi/4 to 0

13) Resolving the limits: $sec ^2 (\pi /4) - 2 log (sec (\pi /4)) - sec ^2 (0) - 2 log (sec (0))$

14) $(\sqrt{2})^2 - 2 log(\sqrt{2}) - (1)^2 - 2 log(1)$

15) $2- \frac{2}{2} log 2 - 1 - 2*0$

16) Hence, the answer is 1- log 2