5
$\begingroup$

Hi I am trying to evaluate the definite integral which has a closed form given by: $$ \mathcal{I}=\int_0^{\pi/4}\frac{x^2\tan x}{\cos^2 x}dx=\frac{\log 2}{2}-\frac{\pi}{4}+\frac{\pi^2}{16}. $$ We can possibly write $y=\cos x$ $$ \mathcal{I}=\int_0^{\pi/4}\frac{x^2\sin x}{\cos^3x}dx=\int_{\frac{1}{\sqrt 2}}^1\frac{(\cos^{-1}y)^2}{y^3}dy $$ however I am now sure how that will help us.
We can also try $u=\tan x, du=\frac{dx}{\cos^2x}$ to obtain $$ \mathcal{I}=\int_0^1 (\tan^{-1}u)^2 u\, du $$ however I am unsure where to go from here. How can we solve $\mathcal{I}$? Thank you

$\endgroup$
  • $\begingroup$ $\tan'x=\dfrac1{\cos^2x}$ $\endgroup$ – Lucian Jun 2 '14 at 15:46
  • $\begingroup$ @Integrals: Seriously, this was again an easy one. :P $\endgroup$ – Pranav Arora Jun 2 '14 at 15:49
  • $\begingroup$ @PranavArora Do you want me to post harder ones? I still have unsolved integrals on math stack, so I am not sure of the level of difficult to post. $\endgroup$ – Jeff Faraci Jun 2 '14 at 15:52
  • $\begingroup$ I can help with slightly harder ones, the really hard ones are unapproachable (for me at least). :P The integrals posted by you which are unsolved are impossible for me, too much for a high school student.:3 Maybe the master solvers on this website can comment on what type of integrals you should post. :) $\endgroup$ – Pranav Arora Jun 2 '14 at 15:57
11
$\begingroup$

$$I=\int_0^{\pi/4} \frac{x^2\tan x}{\cos^2 x}\,dx=\int_0^{\pi/4} x^2\tan x\sec^2 x\,dx$$ Now use integration by parts: $$I=\left(x^2 \frac{\tan^2x}{2}\right|_0^{\pi/4}-\int_0^{\pi/4} x\tan^2 x\,dx=\frac{\pi^2}{32}-\int_0^{\pi/4} x\tan^2 x\,dx$$ Use integration by parts again to evaluate the last integral: $$I=\frac{\pi^2}{32}-\left(x(\tan x-x)\right|_0^{\pi/4}+\int_0^{\pi/4}(\tan x-x)\,dx$$ $$\Rightarrow I=\frac{\pi^2}{32}-\frac{\pi}{4}+\frac{\pi^2}{16}+\frac{\ln 2}{2}-\frac{\pi^2}{32}=\boxed{\dfrac{\ln 2}{2}-\dfrac{\pi}{4}+\dfrac{\pi^2}{16}}$$ $\blacksquare$

$\endgroup$
3
$\begingroup$

$$J=\int x^2\frac{\tan x}{\cos^2x}dx=\int (x^2)\frac{\sin x}{\cos^3x}dx$$

Integrating by Parts,

$$J=x^2\int\frac{\sin x}{\cos^3x}dx-\int\left(\frac{d(x^2)}{dx}\frac{\sin x}{\cos^3x}dx\right)dx$$

Now $\displaystyle\frac{\sin x}{\cos^3x}dx=-\int\frac{d(\cos x)}{\cos^3x}=-\sec^2x+K$

Again, $$\int x\sec^2x\ dx=x\int\sec^2x\ dx-\int\left(\frac{dx}{dx}\int\sec^2x\ dx\right)dx$$

$\endgroup$

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.