Question

How should I solve this?

Answer 1

Question:

Evalute:

$\int\limits_{0}^{ π } \frac{x \: sinx}{1 + sinx} dx$

Solution:

$let \: I =\int\limits_{0}^{π} \frac{xsinx}{1 + sinx} dx ........(1)\\ \\ I = \int\limits_{0}^{π} \frac{( π - x)sinx( π - x) }{1 + sin( π - x} dx$

Note: Using Property

$\int\limits_{0}^{a}f(x)dx = \int\limits_{0}^{a}f( a- x)dx$

I=$\int\limits_{0}^{π} \frac{(π - x)sinx}{1 + sinx} ..............(2)$

Adding (1) and (2) ,

$2I = \int\limits_{0}^{π} \frac{πsinx}{1 + sinx} dx \\ \\ 2I = π \int\limits_{0}^{π} \frac{sinx}{1 + sinx} dx \\ \\ I = \frac{π}{2} \int\limits_{0}^{π} \frac{sinx}{1 + sinx} dx.........(3)$

Now,

$I = \int\limits_{0}^{π} \frac{sinx}{1 + sinx}$

[Add and subtract 1 ]

$I= \int\limits_{0}^{π} \frac{(1 + sinx) - 1}{1 + sinx} dx \\ \\I = \int\limits_{0}^{π}(1 - \frac{1}{1 + sinx } )dx \\ \\ \: \: = \int\limits_{0}^{π}1.dx - \int\limits_{0}^{π} \frac{1 - sinx}{(1 + sinx)(1 - sinx)} dx \\ \\ \: \: = [x]limit_{0}^{π} - \int\limits_{0}^{π} \frac{1 - sinx}{1 - {sin}^{2}x } dx \\ \\ \: \: = (π - 0) - \int\limits_{0}^{π} \frac{1 - sinx}{ {cos}^{2}x } dx \\ \\ \: \: = π - \int\limits_{0}^{π} ({sec}^{2} x - secx \: tanx)dx \\ \\ \: \: = π- [tanx - secx]limit_{0}^{π} \\ \\ \: \: = π - [(tan π - secπ) - (tan0 - sec0)] \\ \\ \: \: = π - [(0 - 1) - (0 - 1)] \\ \\ \: \: = π - 2$

Putting in (3),

$I = \frac{π}{2} (π - 2) \\ \\ \: \: = \frac{ {π}^{2} }{2} - π$