4
$\begingroup$

Consider this integral

$$\int_{0}^{\pi/4}\left[4\cos^2(x)-1\right]\color{red}{\sqrt{\tan x}}\,\mathrm dx\tag1=1$$

An attempt:

Rewrite $(1)$ as

$$\int_{0}^{\pi/4}4\cos^2(x)\sqrt{\tan x}\,\mathrm dx-\int_{0}^{\pi/4}\sqrt{\tan x}\,\mathrm dx=I_1-I_2\tag2$$

For $I_2$ we choose $u=\sqrt{\tan x}\implies {2u\over 1+u^2}dx=dx$, then it becomes

$$\int_{0}^{1}{2u^2\mathrm du\over 1+u^4}\tag3$$

Rearrange $(3)$ in the form of

$$\int_{0}^{1}{1+{1\over u^2}\over u^2+{1\over u^2}}\mathrm du+\int_{0}^{1}{1-{1\over u^2}\over u^2+{1\over u^2}}\mathrm du\tag4$$

Apply completing the square to $(4)$

$$\int_{0}^{1}{1+{1\over u^2}\over \left(u-{1\over u}\right)^2+2}\mathrm du+\int_{0}^{1}{1-{1\over u^2}\over \left(u+{1\over u}\right)^2-2}\mathrm du=I_3+I_4\tag5$$

For $I_3$ setting $v_1=u-{1\over u}$ and for $I_4$ setting $v_2=u+{1\over u}$ then

$$\left.{1\over \sqrt{2}}\tan^{-1}\left({v_1\over \sqrt{2}}\right)\right|_{0}^{\infty}-{1\over \sqrt{8}} \left.\ln{\left({v_2+\sqrt{2}\over v_2-\sqrt{2}}\right)}\right|_{2}^{\infty}={\pi\over \sqrt{8}}+{1\over \sqrt{8}}\ln{(3+2\sqrt{2})}\tag6$$

We can rewrite $I_1$ as

$$I_1=2\int_{0}^{\pi/4}\sqrt{\tan x}\,\mathrm dx+2\int_{0}^{\pi/4}\cos(2x)\sqrt{\tan x}\,\mathrm dx\tag7$$

$$I_1=2\cdot(6)+2\int_{0}^{\pi/4}\cos(2x)\sqrt{\tan x}\,\mathrm dx\tag8$$

$(8)$ we can apply integration by part, I am sure it would be lengthy.

How else can we tackle $(1)$ in another less lengthy way?

$\endgroup$
13
+50
$\begingroup$

What has been seen cannot be unseen: $$ \begin{aligned} \int (4\cos^2x-1)\sqrt{\tan x}\,dx&=\int (1+2\cos 2x)\sqrt{\tan x}\,dx\\ &=\int\sin 2x\frac{1}{2\sqrt{\tan x}}\frac{1}{\cos^2 x}+2\cos 2x\sqrt{\tan x}\,dx\\ &=\int \sin2x\frac{d}{dx}\sqrt{\tan x}+\Bigl(\frac{d}{dx}\sin 2x\Bigr)\sqrt{\tan x}\,dx\\ &=\int \frac{d}{dx}\Bigl(\sin 2x\sqrt{\tan x}\Bigr)\,dx\\ &=\sin 2x\sqrt{\tan x}+C. \end{aligned} $$ Then just insert limits, and you will get

$$\int_0^{\pi/4}\bigl(4\cos^2x-1\bigr)\sqrt{\tan x}\,dx=1.$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Excellent technique $\endgroup$ – gymbvghjkgkjkhgfkl Apr 9 '17 at 14:34
  • 2
    $\begingroup$ Amazing. I wonder how one comes up with the second step though... $\endgroup$ – rubik Apr 10 '17 at 6:05
5
$\begingroup$

I don't see an easy approach to tackle your last integral : you can apply the method I will show just after this block of text, but it is faster to apply it directly to the initial integral.

Here is the other approach : $$I=\int_{0}^{\pi/4}\left[4\cos^2(x)-1\right]{\sqrt{\tan x}}\mathrm dx={\displaystyle\int}_0^{\pi/4}\class{steps-node}{\cssId{steps-node-1}{\sec^2\left(x\right)}}\class{steps-node}{\cssId{steps-node-2}{\left(-\dfrac{\sqrt{\tan\left(x\right)}\left(\tan^2\left(x\right)-3\right)}{\left(\tan^2\left(x\right)+1\right)^2}\right)}}\,\mathrm{d}x$$ By using the identities $\cos(x)=\frac{1}{\sec(x)}$ and $\sec^2(x)=1+\tan^2(x)$.

Now you can substitute $u=\tan(x)$ and $du=\sec^2(x) dx$. $$I=-{\displaystyle\int}_0^1\dfrac{\sqrt{u}\left(u^2-3\right)}{\left(u^2+1\right)^2}\,\mathrm{d}u$$ Now you can substitute $v=\sqrt u$ and $dv=\frac{du}{2\sqrt u}$. $$I=\class{steps-node}{\cssId{steps-node-3}{-2}}{\displaystyle\int}_0^1\dfrac{v^2\left(v^4-3\right)}{\left(v^4+1\right)^2}\,\mathrm{d}v$$ Now you can do a long division and finish.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Nice and neat (+1) $\endgroup$ – gymbvghjkgkjkhgfkl Apr 9 '17 at 8:16
3
$\begingroup$

$\displaystyle J=\int_{0}^{\pi/4}\left[4\cos^2(x)-1\right]{\sqrt{\tan x}}\,\mathrm dx$

Perform the change of variable $y=\dfrac{\pi}{2}-x$,

$\displaystyle J=\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\dfrac{4(\sin x)^2-1}{\sqrt{\tan x}}dx$

Perform the change of variable $y=\sqrt{\tan x}$,

$\begin{align} J&=\int_{1}^{\infty} \dfrac{\tfrac{4}{1+\frac{1}{x^4}}-1}{x}\times \dfrac{2x}{1+x^4}dx\\ &=2\int_{1}^{\infty} \dfrac{3x^4-1}{(1+x^4)^2}dx\\ &=2\left[\dfrac{-x}{1+x^4}\right]_{1}^{\infty}\\ &=2\times \left(0+\dfrac{1}{2}\right)\\ &=\boxed{1} \end{align}$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.