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## Homework Statement

п/2

∫dx/(2+sinx)

^{2}

0

## Homework Equations

## The Attempt at a Solution

п/2

∫dx/(2 + sinx)

^{2}=

0

п/2

∫dx/(4 + 4sinx + (sinx)

^{2})

0

substitude t=tg x/2 => x=2arctgx

dx=2/(1+t

^{2}) dt

sinx=2t/(1+t

^{2})

t=tgx/2 =>

1

∫(1/(4 + 4*(2t/(1+t

^{2})) +(2t/(1+t

^{2}))

^{2})) * (2/(1+t

^{2}))dt =

0

1

∫( ((1+t

^{2})

^{2})/(2* (2(1+t

^{2})

^{2}+4t(1+t

^{2}) +4t

^{2}) ) ) * (2/(1+t

^{2}))dt [

0

1

=∫(1+t

^{2})/(2+4t

^{2}+2t

^{4}+4t+4t

^{3}+4t

^{2}) dt

0

= 1/2 ∫(1+t

^{2})/(t

^{4}+2t

^{3}+4t

^{2}+2t+1) dt

And here it is divided into two integrals, but i can't come to an answer