Question

integrate ------ 0 to π/2
{(x+sinx)/(1+cosx)}

Answer 1

✔️✔️divide it in 2  parts and then solve.

⭕️1st part use Integration by parts.

⭕️2nd part use 1+cosx = t

✔️✔️In the equation [1] in above attachment use the formula which is given in "blue ink"

✴️If you're further having any doubts ask in comment section✴️

There's a correction in the equation <A>

note that in 3rd attachment!!!!!



$\huge\bf{\boxed {\red{\mathfrak{thank \: you :)}}}}$

Answer 2

Answer: π/2

Step-by-step explanation:

Given,

$I=\int\limits^\frac{\pi}{2}_0{\frac{x+\sin x}{1+\cos x}}\,dx\\\;\\I=\int\limits^\frac{\pi}{2}_0{\frac{x+2\sin\frac{x}{2}.\cos\frac{x}{2}}{1+2\cos^2\frac{x}{2}-1}}\,dx\\\;$

$I=\int\limits^\frac{\pi}{2}_0{\frac{x+2\sin\frac{x}{2}.\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}}\,dx\\\;$

$I=\int\limits^\frac{\pi}{2}_0{\frac{x}{2\cos^2\frac{x}{2}}}\,dx+\int\limits^\frac{\pi}{2}_0{\frac{2\sin\frac{x}{2}.\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}}\,dx\\\;$

$I=\frac{1}{2}\int\limits^\frac{\pi}{2}_0{x\sec^2\frac{x}{2}}\,dx+\int\limits^\frac{\pi}{2}_0{\frac{2\sin\frac{x}{2}}{2\cos\frac{x}{2}}}\,dx\\\;$

$I=\frac{1}{2}\int\limits^\frac{\pi}{2}_0{x\sec^2\frac{x}{2}}\,dx+\int\limits^\frac{\pi}{2}_0{\tan\frac{x}{2}}\,dx\\\;$

Using  Int. by  Parts in first  term.

$I=\frac{1}{2}[x\int\limits^\frac{\pi}{2}_0{\sec^2\frac{x}{2}}\,dx-\int\limits^\frac{\pi}{2}_0{\frac{dx}{dx}}.\int\limits^\frac{\pi}{2}_0{\sec^2\frac{x}{2}\,dx)\,dx}]+\int\limits^\frac{\pi}{2}_0{\tan\frac{x}{2}}\,dx\\\;$

$I=[x\tan\frac{x}{2}]\limits^\frac{\pi}{2}_0-\frac{2}{2}\int\limits^\frac{\pi}{2}_0{\tan\frac{x}{2}}\,dx+\int\limits^\frac{\pi}{2}_0{\tan\frac{x}{2}}\,dx\\\;$

$I=[x\tan\frac{x}{2}]\limits^\frac{\pi}{2}_0-\int\limits^\frac{\pi}{2}_0\tan\frac{x}{2}\,dx+\int\limits^\frac{\pi}{2}_0{\tan\frac{x}{2}}\,dx\\\;\\I=[x\tan\frac{x}{2}]\limits^\frac{\pi}{2}_0\\\;\\\text{Putting Limits}$

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