Question

π/2
∫ sin²x/(1+cosx)² dx ,Evaluate it.
0

Answer 1

please see the attachment

Answer 2

Answer:

$2-\frac{\pi}{2}$

Step-by-step explanation:

In the given question,

$\int\limits^\frac{\pi}{2}_0 {\frac{sin^{2}x}{(1+cosx)^{2}}} \, dx$

Now on putting the value of,

$sin^{2}x=1-cos^{2}x$ we get,

$\int\limits^\frac{\pi}{2}_0 {\frac{sin^{2}x}{(1+cosx)^{2}}} \, dx =\int\limits^\frac{\pi}{2}_0 {\frac{1-cos^{2}x}{(1+cosx)^{2}}} \, dx$

So, on doing that,

$\int\limits^\frac{\pi}{2}_0 {\frac{1-cos^{2}x}{(1+cosx)^{2}}} \, dx =\int\limits^\frac{\pi}{2}_0 {\frac{(1+cosx)(1-cosx)}{(1+cosx)^{2}}} \, dx$

Now we know that,

$1-cosx=sin^{2}\frac{x}{2}+cos^{2}\frac{x}{2}-cos^{2}\frac{x}{2}+sin^{2}\frac{x}{2}\\=2sin^{2}\frac{x}{2}\\and,\\1+cosx=sin^{2}\frac{x}{2}+cos^{2}\frac{x}{2}+cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}\\=2cos^{2}\frac{x}{2}$

So, on putting the values of above in the equation we get,

$=\int\limits^\frac{\pi}{2}_0 {\frac{1-cosx}{(1+cosx)}} \, dx \\=\int\limits^\frac{\pi}{2}_0 {\frac{2sin^{2}\frac{x}{2}}{2cos^{2}\frac{x}{2}}} \, dx \\=\int\limits^\frac{\pi}{2}_0 {tan^{2}\frac{x}{2}} \, dx\\ =\int\limits^\frac{\pi}{2}_0( {sec^{2}\frac{x}{2}-1}) \, dx\\=[{2tan\frac{x}{2}-x}]^{\frac{\pi}{2}}_{0}$

Therefore, on putting the limits we get,

$(2tan\frac{\pi}{4})-\frac{\pi}{2}-(0-0)=2-\frac{\pi}{2}$

Therefore, the value of the integration given in  the equation is given by,

$(2tan\frac{\pi}{4})-\frac{\pi}{2}-(0-0)=2-\frac{\pi}{2}$