What is antiderivative of function f(x)=((sinx)raised to square+sinx)/(1+sinx+cosx)?calculate for limits x=0 and x=pi/2

Wednesday, June 24, 2015

What is antiderivative of function f(x)=((sinx)raised to square+sinx)/(1+sinx+cosx)?calculate for limits x=0 and x=pi/2

First notice that

(1+cosx+sinx)^2=cos^2x+2cosxsinx+2cosx+sin^2x+2sinx+
1

$\displaystyle={{\cos}}^{{2}}{x}+{{\sin}}^{{2}}{x}+{2}{\cos{{x}}}{\sin{{x}}}+{2}{\cos{{x}}}+{2}{\sin{{x}}}+{1}={1}+{2}{\cos{{x}}}{\sin{{x}}}+{2}{\cos{{x}}}+{2}{\sin{{x}}}+{1}$

$\displaystyle={2}{\left({1}+{\cos{{x}}}{\sin{{x}}}+{\sin{{x}}}+{\cos{{x}}}\right)}={2}{\left({1}+{\cos{{x}}}\right)}{\left({1}+{\sin{{x}}}\right)}$

Now we multiply
$\displaystyle\frac{{{\sin{{x}}}{\left({\sin{{x}}}+{1}\right)}}}{{{1}+{\sin{{x}}}+{\cos{{x}}}}}\cdot\frac{{{1}+{\sin{{x}}}+{\cos{{x}}}}}{{{1}+{\sin{{x}}}+{\cos{{x}}}}}$

We
get

$\displaystyle{\sin{{x}}}{\left({1}+{\sin{{x}}}\right)}\frac{{{1}+{\sin{{x}}}+{\cos{{x}}}}}{{{\left({1}+{\sin{{x}}}+{\cos{{x}}}\right)}}^{{2}}}$
=sinx(1+sinx)

$\displaystyle\frac{{{1}+{\sin{{x}}}+{\cos{{x}}}}}{{{2}{\left({1}+{\cos{{x}}}\right)}{\left({1}+{\sin{{x}}}\right)}}}$

$\displaystyle={\sin{{x}}}\frac{{{1}+{\sin{{x}}}+{\cos{{x}}}}}{{{2}{\left({1}+{\cos{{x}}}\right)}}}$

$\displaystyle={\sin{{x}}}\frac{{{1}+{\cos{{x}}}}}{{{2}{\left({1}+{\cos{{x}}}\right)}}}+{{\sin}}^{{2}}\frac{{x}}{{{2}{\left({1}+{\cos{{x}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\sin{{x}}}+\frac{{{1}-{{\cos}}^{{2}}{x}}}{{{2}{\left({1}+{\cos{{x}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\sin{{x}}}+\frac{{{\left({1}+{\cos{{x}}}\right)}{\left({1}-{\cos{{x}}}\right)}}}{{{2}{\left({1}+{\cos{{x}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\sin{{x}}}+\frac{{{1}-{\cos{{x}}}}}{{2}}$

$\displaystyle=\frac{{1}}{{2}}{\sin{{x}}}+\frac{{1}}{{2}}-\frac{{1}}{{2}}{\cos{{x}}}$

The antiderivative of $\displaystyle\frac{{1}}{{2}}{\sin{{x}}}+\frac{{1}}{{2}}-\frac{{1}}{{2}}{\cos{{x}}}$
is

$\displaystyle-\frac{{1}}{{2}}{\cos{{x}}}+\frac{{1}}{{2}}{x}-\frac{{1}}{{2}}{\sin{{x}}}$
So our answer
is

$\displaystyle\frac{{1}}{{2}}{x}-\frac{{1}}{{2}}{\cos{{x}}}-\frac{{1}}{{2}}{\sin{{x}}}+{C}$

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}$
(sin^2x+sinx)/(1+cosx+sinx)dx

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{\pi}{{2}}\right)}-\frac{{1}}{{2}}\frac{{\cos{\pi}}}{{2}}-\frac{{1}}{{2}}\frac{{\sin{\pi}}}{{2}}$
- (1/2(0) - 1/2cos0-1/2sin0)

$\displaystyle=\frac{\pi}{{4}}-\frac{{1}}{{2}}{\left({0}\right)}-\frac{{1}}{{2}}{\left({1}\right)}-{\left({0}\right.}$
- 1/2(1) - 1/2(0))

$\displaystyle=\frac{\pi}{{4}}-\frac{{1}}{{2}}+\frac{{1}}{{2}}=\frac{\pi}{{4}}$ Which is
our answer

Help Notes

Wednesday, June 24, 2015