Question Number 76716 by peter frank last updated on 29/Dec/19

Commented by abdomathmax last updated on 30/Dec/19

Commented by peter frank last updated on 30/Dec/19

Commented by Tony Lin last updated on 30/Dec/19
 =∫_0 ^π (π−θ)f[sinθ]dθ =π∫_0 ^π f(sinθ)dθ−∫_0 ^π θf(sinθ)dθ =π∫_0 ^π f(sinx)dx−∫_0 ^π xf(sinx)dx ⇒∫_0 ^π xf(sinx)dx=(π/2)∫_0 ^π f(sinx)dx ∫_0 ^π ((xsinx)/(1+cos^2 x))dx =(π/2)∫_0 ^π ((sinx)/(1+cos^2 x))dx let cosx=t, dt=−sinxdx (π/2)∫_(−1) ^1 (1/(1+t^2 ))dt =πtan^(−1) t∣_0 ^1 =(π^2 /4)](https://www.tinkutara.com/question/Q76741.png)
Commented by peter frank last updated on 30/Dec/19

Answered by Kunal12588 last updated on 30/Dec/19
![I=∫_0 ^π ((x sin x)/(1+cos^2 x))dx ⇒I=∫_0 ^π (((π−x) sin (π−x))/(1+cos^2 (π−x)))dx=∫_0 ^π (((π−x)sin x)/(1+cos^2 x))dx [using ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx] adding red ones 2I= π∫_0 ^π ((sin x)/(1+cos^2 x))dx (((t=cos x⇒dt=−sin x dx)),((x→0⇒t→1 & x→π⇒t→−1)) ) 2I=−π∫_1 ^(−1) (dt/(1+t^2 ))=π∫_(−1) ^1 (dt/(1+t^2 ))=2π∫_0 ^1 (dt/(1+t^2 )) [using −∫_b ^a f(x)dx=∫_a ^b f(x)dx] [and ∫_(−a) ^( a) f(x) dx = { ((∫_0 ^( 2a) f(x) dx ; if f(x) is even function)),((0 ; if f(x) is odd function)) :}] ⇒I=π[tan^(−1) t]_0 ^1 =π[(π/4)−0]=(π^2 /4) ∴∫_0 ^π ((x sin x)/(1+cos^2 x))dx=(π^2 /4) QED](https://www.tinkutara.com/question/Q76754.png)
Commented by peter frank last updated on 30/Dec/19
