Trigonometric Substitution in $\int _0^{\pi/2}{\frac{ x\cos x}{ 1+\sin^2 x} dx }$

Question

Evaluate $$ \int _{ 0 }^{ \pi /2 }{ \frac { x\cos { (x) } }{ 1+\sin ^{ 2 }{ x } } \ \mathrm{d}x } $$ $$$$ The solution was suggested like this:$$$$ SOLUTION: First of all its, quite obvious to have substitution $ \sin(x) \rightarrow x $ $$ I = \int_{0}^{1} \frac{\arcsin(x)}{1+x^2} \ \mathrm{d}x$$ Now using integration by parts, $$ I = \frac{\pi^2}{8} - \int_{0}^{1} \frac{\arctan(x)}{\sqrt{1-x^2}} \ \mathrm{d}x$$ Could someone please explain these two steps to me? For example, how do we get $\arcsin(x)$ in the numerator? Thanks a lot!

Answer 1

Since setting $\sin x=u$ gives you $$x=\arcsin u,\ \ \ \mathrm{d}u=\cos x \ \mathrm{d}x,$$ you have $$\begin{align}\int_{0}^{\pi/2}\frac{x\cos x}{1+\sin^2x}\ \mathrm{d}x&=\int_{0}^{1}\frac{\arcsin u}{1+u^2}\ \mathrm{d}u\\&=\int_{0}^{1}(\arctan u)'\arcsin u\ \ \mathrm{d}u\\&=[\arctan u\arcsin u]_{0}^{1}-\int_{0}^{1}\frac{\arctan u}{\sqrt{1-u^2}}\ \mathrm{d}u\\&=\frac{\pi^2}{8}-\int_{0}^{1}\frac{\arctan u}{\sqrt{1-u^2}}\ \mathrm{d}u\end{align}$$

Answer 2

It's a change of variables. If $y=\sin(x)$, $x=\arcsin(y)$ and $dy=\cos(x)dx$, so $x\cos(x)dx=\arcsin(y)dy$. Then you also have to change the bounds of the integral.

Knowing that :

• the derivative of $x\mapsto \arcsin(x)$ is $x\mapsto\frac{1}{1-\sqrt{x^2}}$

• the derivative of $x\mapsto \arctan(x)$ is $x\mapsto\frac{1}{1+x^2}$

the second part is simple if you know how to integrate by parts.

Answer 3

You may obtain

$$ \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx=\frac12 \log^2 (1+\sqrt{2}) . \tag1 $$

Proof. First observe that $$\begin{align} \int_0^\pi \frac{x\cos x}{1+\sin^2 x}\:dx &= \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx + \int_{\pi/2}^{\pi} \frac{x\cos x}{1+\sin^2 x}\:dx \\ & = \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \int_0^{\pi/2} \frac{(x+\pi/2)\sin x}{1+\cos^2 x}\:dx \\ &= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \pi\int_0^{\pi/2} \frac{\cos x}{1+\sin^2 x}\:dx \\ &= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \pi \left[\arctan u\right]_0^1\\ &= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \frac{\pi^2}{4}. \end{align} $$ Then evaluate $$ \int_0^{\pi}\frac{x \cos{x}}{1+\sin^2{x}} {\rm d}x $$ using for example this approach.

Answer 4

With a few transformations we can obtain

$$\begin{align*} & \int_0^{\tfrac\pi2} \frac{x \cos x}{1 + \sin^2x} \, dx \\ &= 4 \int_1^\infty \frac{y \arctan y}{1+y^4} \, dy - \pi \int_1^\infty \frac y{1+y^4} \, dy \tag1 \\ &= 2 \int_1^\infty \frac{\arctan \sqrt z}{1+z^2} \, dz - \frac\pi2 \int_1^\infty \frac{dz}{1+z^2} \tag2 \\ &= 2\left(\frac{3\pi^2}{16} - \frac12 \int_1^\infty \frac{\arctan z}{\sqrt z\,(1+z)}\,dz\right) - \frac{\pi^2}8 \tag3 \\ &= \frac{\pi^2}4 - 2 \int_1^\infty \frac{\arctan w^2}{1+w^2} \, dw \tag4 \\ &= \frac{\pi^2}4 - 2 \left(\frac{\pi^2}8 - \int_0^1 \frac{\arctan v^2}{1+v^2} \, dv\right) \tag5 \\ &= 2 \int_0^1 \frac{\arctan v^2}{1+v^2} \, dv = \boxed{\frac12\ln^2\left(1+\sqrt 2\right)} \end{align*}$$

The last integral is done here.

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• $(1)$ Substitute $y=\sec x+\tan x \iff x=-2\arctan\dfrac{1-y}{1+y}=2\arctan y-\dfrac\pi2$
• $(2)$ Substitute $z=\sqrt y$
• $(3)$ Integrate by parts
• $(4)$ Substitute $w=\sqrt z$
• $(5)$ Substitute $v=\dfrac1w$ and make use of $\arctan \dfrac1x+\arctan x=\dfrac\pi2$ for $x>0$