Calculate the integral using another integral

Question

Need help with this integration:

Let $$A = \int_0^\pi \frac{\cos x}{(x+2)^2}dx$$ Compute $$\int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{x+1}dx$$

In terms of $A$.

I tried to do some algebraic manipulations but either get something rather complicated or just dead ends. Any help would be great!

Answer 1

In the original integral change variables $x = u/2$ $$ I=\int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{x+1}dx = \int_0^{\pi} \frac{\sin(u)}{2(u+2)} \mathrm{d}u $$ Now integrate by parts: $$ I = \left[ -\frac{\cos(u)}{2(u+2)} \right]_{u=0}^{u = \pi} - \frac{1}{2} \int_0^\pi \frac{\cos(u)}{(u+2)^2} \mathrm{d}u = \frac{1}{4} + \frac{1}{2(\pi+2)} - \frac{1}{2} A $$

Answer 2

Hint: In the second integral, let $x=t/2$. Then $dx=dt/2$, and the second integral is $$\int_0^\pi \frac{\sin(t/2)\cos(t/2)}{t+2}\,dt.$$ Note that $\sin(t/2)\cos(t/2)$ is $\frac{1}{2}\sin t$.

Now use integration by parts. Let $u=\frac{1}{t+2}$ and $dv=\frac{1}{2}\sin t\,dt$.

Answer 3

$$ \begin{align} \int_0^{\frac\pi2}\frac{\sin(x)\cos(x)}{x+1}\mathrm{d}x &=\frac12\int_0^{\frac\pi2}\frac{\sin(2x)}{x+1}\mathrm{d}x\tag{1}\\ &=\frac12\int_0^\pi\frac{\sin(x)}{x+2}\mathrm{d}x\tag{2}\\ &=-\frac12\int_0^\pi\frac1{x+2}\mathrm{d}\cos(x)\tag{3}\\ &=-\frac12\left[\frac{\cos(x)}{x+2}\right]_0^\pi-\frac12\int_0^\pi\frac{\cos(x)} {(x+2)^2}\mathrm{d}x\tag{4}\\ &=\frac12\left(\frac1{\pi+2}+\frac12\right)-\frac12A\tag{5} \end{align} $$ Explanation:
$(1)$: use $\sin(2x)=2\sin(x)\cos(x)$
$(2)$: substitute $x\mapsto x/2$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: plug in limits and use $A$

Answer 4

$$\int_0^{\pi/2} \frac{\sin(x)\cos(x)}{1+x}\,{\rm d}x$$

$$\int_0^{\pi/2} \frac{\sin(2x)}{2(1+x)}\,dx$$

With the substituition $u=2x$

$$\frac12\int_0^{\pi}\frac{\sin(u)}{u+2}\,du$$

Integrating by parts,

$$\frac12\left(\left[-\frac{\cos(u)}{u+2}\right]_0^\pi - A\right)$$