Compute an integral in terms of a given integral

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

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

One comment

September 3, 2019 at 11:19 am

Anonymous says:

As an alternative it’s also possible to use integration by parts with:

(1) $\begin{equation*} u = \frac{\sqrt{a + bx}}{x} \implies du = -\frac{a+bx}{2x^2\sqrt{a+bx}} \end{equation*}$

and

(2) $\begin{equation*} dv = dx \implies v = x \end{equation*}$

Stumbling Robot

Compute an integral in terms of a given integral

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