Verified answer

As an indefinite integral, this one is too

advanced. it cannot be expressed in terms

of elementary transcendental functions.

For more about this integral, see :

http://www.wolframalpha.com/input/?i=int+%28+%28+x...

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To start with, you may consider

dv = ( cos x dx ) / ( 1 + sin² x )

so that

v = ∫ [ 1 / ( 1 + sin² x ) ]· cos x dx = tanֿ¹ ( sin x ) or - cotֿ¹ ( sin x ).

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To Hello :

This is not an indefinite integral.

It is a Definite Integral from x = -π

to x = π.

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Being an odd function, its value is 0.

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This is definite integral

using integration by parts we will get

u=x

du=dx

dv=[cosx/(1+sin^2x)]dx...........1

integrate dv now

take t=sinx

differentiating t with respect to x you will get

dt=cosxdx

substitute dt insturd of cosxdx

as dt

so int(dt/(1+t^2))

v=tan^-1(t)+c

substitute back t=sinx

v=tan^-1(sinx)+c

then substitute all the above values in this equation

uv-int(vdu)

[xcosx/(1+sin^2x)]-int(tan^-1(sinx) dx)