Prove that \int_0^a f(x)dx = \int_0^a f(a-x)dx, hence evaluate \int_0^\pi \frac{x\sin x}{1 + cos^2 x}dx.

 

 

 

 
 
 
 
 

Answers (1)

To prove\int_0^a f(x)dx = \int_0^a f(a-x)dx

Let x = a -t

     dx = -dt

    \\x = 0, t = a \\x = a, t = 0

\int_0^a f(x)dx = - \int_0^a f(a-t)dt

\int_0^a f(x)dx = -\int^0_a f(a-t)dt

                            [ taking minus sign to invert the integral \int_b^a f(x)dx = -\int^b_a f(x)dx]

\Rightarrow \int_0^a f(x)dx = \int^a_0 f(a-x)dx

Hence Proved

\int_0^\pi \frac{x\sin x}{1 + cos^2 x}dx.

 

                  \int_0^a f(x)dx = \int_0^a f(a-x)dx

I = \int_0^\pi \frac{(\pi -x)\sin (\pi -x)}{1 + cos^2 (\pi-x)}dx

    =\int_0^\pi \frac{(\pi -x)\sin x}{1 + (-cos x)^2}dx

    =\int_0^\pi \frac{(\pi -x)\sin x}{1 + cos^2 x}dx

    =\int_0^\pi \frac{\pi\sin x}{1 + cos^2 x}dx - \int_0^\pi \frac{x\sin x}{1 + cos^2 x}dx

I =\int_0^\pi \frac{\pi\sin x}{1 + cos^2 x}dx -I

2I =\pi\int_0^\pi \frac{\sin x}{1 + cos^2 x}dx

\\\cos x = t \\-\sin x = dt \\ \sin x = - dt \\ x\rightarrow 0 \quad t = 1 \\ x = \pi \quad t = -1

2I =\pi\int_{1}^{-1} \frac{-dt}{1 + t^2}

I =\frac{1}{2}\pi\int^{1}_{-1} \frac{dt}{1 + t^2}

I =\frac{\pi}{2}\int^{1}_{-1} \frac{dt}{1 + t^2}

I =\frac{\pi}{2}\left[\tan^{-1}t \right ]_{t=-1}^{t = 1}

I =\frac{\pi}{2}\left[\tan^{-1}1 - \tan^{-1}(-1) \right ]

I =\frac{\pi}{2}\left[\frac{\pi}{4} - \left(-\frac{\pi}{4} \right ) \right ]

    = \frac{\pi^2}{4}

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