Xsinx definite integral

Xsinx definite integral

There are no upper and lower bounds, so we can only find indefinite integrals,
∫xsinxdx=sinx-xcosx+C
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The integral of (x + SiNx) DX / 1 + cosx

(x + SiNx) DX / 1 + cosx
＝(x+sinx)(1-cosx)dx/(1+cosx)(1-cosx)
=(x-xcosx+sinx-sinxcosx)dx/sin^2x
Expand separately
Can you do it? Maybe you can do it

Integral (0, pie) xsinx / (1 + (cosx) ^ 2) DX
The integral range is from zero to pie

let f(x) = xsinx/(1+(cosx)^2f(-x) = f(x) ie f(x) is even function∫(0,π）xsinx/(1+(cosx)^2) dx = ∫(-π,0）xsinx/(1+(cosx)^2) dxI= ∫(0,π）xsinx/(1+(cosx)^2) dx (1)let y =(π-x)dy = dxx=0,y=πx=π ,...

(1 + x ^ 2) ^ 0.5dx for definite integral
It's an indefinite integral

The answer is the same as above,
∫√(1+x^2)dx x=tanu √1+x^2=secu
∫secudtanu
=tanusecu-∫tanudsecu
=tanusecu-∫secutanu^2du
=tanusecu-∫secu(secu^2-1)du
=tanusecu-∫secu^3du+∫secudu
=tanusecu-∫secudtanu+∫du/cosu
2∫secudtanu=tanusecu+∫du/cosu
∫secutanu=(1/2)tanusecu+(1/2)ln|secu+tanu|+C
∫√(1+x^2)dx
=(1/2) x√1+x^2) +(1/2)ln|x+√(1+x^2)|+C