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Compute an integral in terms of a given integral

by RoRi

Let

    \[ A = \int_0^{\pi} \frac{\cos x}{(x+2)^2} \, dx. \]

Compute the following integral (in terms of A):

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

Starting with the expression for A we integrate by parts using

    \begin{align*} u &= \cos x & du &= - \sin x \, dx \\ dv &= \frac{1}{(x+2)^2} \, dx & v &= -\frac{1}{x+2}. \end{align*}

Therefore, we have

    \begin{align*} A = \int_0^{\pi} \frac{\cos x}{(x+2)^2} \, dx &= \left. -\frac{\cos x}{x+2} \right|_0^{\pi} - \int_0^{\pi} \frac{\sin x}{x+2} \, dx \\ &= \left( \frac{1}{\pi+2} + \frac{1}{2} \right) - \int_0^{\pi} \frac{\sin x}{x+2} \, dx \\ &= \left( \frac{1}{2} + \frac{1}{\pi+2} \right) - 2 \int_0^{\frac{\pi}{2}} \frac{\sin (2x)}{2x+2} \, dx \intertext{using the expansion/contraction property of the integral, Theorem 1.19 on page 81 of Apostol,} &= \left( \frac{1}{2} + \frac{1}{\pi + 2} \right) - 2 \int_0^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{2 (x+1)} \, dx \\ &= \left( \frac{1}{2} + \frac{1}{\pi + 2} \right) - 2 \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{x+1} \, dx. \end{align*}

Solving for the integral in terms of A we have

    \[ \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{x+2} \, dx = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{\pi + 2} - A \right). \]

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