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Prove another identity of the integral of some trig functions

  1. Prove the following identity:

        \[ \int_0^{\pi} x f(\sin x) \, dx = \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx. \]

  2. Using part (a) deduce the formula

        \[ \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \pi \int_0^1 \frac{dx}{1+x^2}. \]


  1. Proof. Following the hint, we make the substitution

        \[ u = \pi - x \qquad du = -dx. \]

    So we then have

        \begin{align*}  \int_0^{\pi} x f(\sin x) \, dx &= - \int_{u(0)}^{u(\pi)} (\pi - u) f(\sin (\pi - u)) \, du \\  &= -\int_{\pi}^0 (\pi - u) f(\sin u) \, du \\  &= \int_0^{\pi} (\pi - u) f(\sin u) \, du \\  &= \int_0^{\pi} \pi f(\sin u) \, du - \int_0^{\pi} u f (\sin u) \, du  \end{align*}

    Here, we change the the name of the variable of integration from u to x. (We can always rename the variable of integration since integrating f(t) \, dt is the same as integrating f(x) \, dx, for example.) So this means we have

        \begin{align*}  &&\int_0^{\pi} x f(\sin x) \, dx &= \int_0^{\pi} \pi f(\sin x) \, dx - \int_0^{\pi} x f (\sin x) \, dx \\  \implies && \int_0^{\pi} x f(\sin x) \, dx &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx. \qquad \blacksquare \end{align*}

  2. Now to deduce the requested formula, first we use the trig identity \cos^2 x + \sin^2x = 1, which implies 1 + \cos^2 x = 2 - \sin^2 x, to rewrite the integral

        \begin{align*}  \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx &= \int_0^{\pi} \frac{x \sin x}{2 - \sin^2 x} \, dx \\  &= \int_0^{\pi} x f( \sin x) \, dx \qquad \text{where} \qquad f(x) = \frac{x}{2-x^2}. \end{align*}

    Then, using the formula we established in part (a) we have

        \begin{align*}  \int_0^{\pi} x f(\sin x) \, dx &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx \\  &= \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{2-\sin^2 x} \, dx \\  &= \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \, dx. \end{align*}

    Now, we use the substitution method, letting

        \[ u = \cos x \qquad du = -\sin x \, dx. \]

    So we obtain,

        \begin{align*}  \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \,dx &= -\frac{\pi}{2} \int_1^{-1} \frac{1}{1+u^2} \, du \\  &= \frac{\pi}{2} \int_{-1}^1 \frac{1}{1+x^2} \, dx &(\text{renaming variable of integration}) \\  &= \pi \int_0^1 \frac{1}{1+x^2} \, dx. \end{align*}

    The last equality follows since \frac{1}{1+(-x)^2} = \frac{1}{1+x^2} so that this is an even function. Hence (by a previous exercise) the integral from -1 to 1 is twice the integral from 0 to 1. \qquad \blacksquare

Prove expansion/contraction and translation invariance of interval of integration using method of substitution

Use the method of substitution to prove invariance under translation (Theorem 1.18 on page 81 of Apostol) and to prove expansion or contraction of the interval of integration (Theorem 1.19 on page 81 of Apostol).


Theorem: (Invariance Under Translation) For a function f integrable on an interval [a,b] and for every c \in \mathbb{R} we have

    \[ \int_a^b f(x) \, dx = \int_{a+c}^{b+c} f(x-c) \, dx. \]

Proof. If P is a primitive of f, then

    \[ \int_a^b f(x) \, dx = P(b) - P(a). \]

Let

    \[ u = g(x) = x-c \quad \implies \quad du = g'(x) \, dx = dx. \]

So,

    \begin{align*}  \int_{a+c}^{b+c} f(x-c) \, dx &= \int_{a+c}^{b+c} f(g(x)) g'(x) \, dx \\  &= P(g(b+c)) - P(g(a+c)) \\  &= P(b) - P(a). \end{align*}

Hence, we indeed have

    \[ \int_a^b f(x) \,dx = \int_{a+c}^{b+c} f(x-c) \, dx. \qquad \blacksquare \]

Theorem: (Expansion or Contraction of the Interval of Integration) For a function f integrable on an interval [a,b] and for every k \in \mathbb{R} with k \neq 0,

    \[ \int_a^b f(x) \, dx = \frac{1}{k} \int_{ka}^{kb} f \left(\frac{x}{k} \right) \, dx. \]

Proof. Let

    \[ u = g(x) = \frac{x}{k} \quad \implies \quad du = g'(x) \, dx = \frac{1}{k} \, dx. \]

Then we have

    \begin{align*}  \frac{1}{k} \int_{ka}^{kb} f \left( \frac{x}{k} \right) \, dx &= \int_{ka}^{kb} f(g(x)) g'(x) \, dx \\  &= \int_{g(ka)}^{g(kb)} f(u) \, du \\  &= P(g(b)) - P(g(a)) \\  &= P(b) - P(a). \end{align*}

Thus, we indeed have

    \[ \int_a^b f(x) \, dx = \frac{1}{k} \int_{ka}^{kb} f \left( \frac{x}{k} \right) \, dx. \qquad \blacksquare \]