13
$\begingroup$

I have$$\int\limits_{0}^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}$$ Tangent half-angle substitution gives a fourth-degree polynomial in the denominator that is difficult to factor.

$\endgroup$
  • $\begingroup$ Isn't this a duplicate? $\endgroup$ – Nikunj Mar 24 '16 at 23:01
  • $\begingroup$ @Nikunj You shall link the "original" question then. $\endgroup$ – Turing Mar 24 '16 at 23:08
  • $\begingroup$ The indefinite integral is discussed here: math.stackexchange.com/q/684053/215011 $\endgroup$ – grand_chat Mar 24 '16 at 23:09
10
$\begingroup$

$$I = \int_{0}^{\pi/2}\frac{dx}{\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\left(1-\frac{1}{2}\sin(2x)\right)}=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sqrt{2}\cos(x)\left(1-\frac{1}{2}\cos(2x)\right)} $$ hence, through the substitution $x=\arcsin t$: $$ I = \sqrt{2}\int_{0}^{\pi/4}\frac{dx}{\cos(x)\left(\frac{3}{2}-\cos^2 x\right)}=\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{dt}{(1-t^2)\left(\frac{1}{2}+t^2\right)}$$ and the last integral is perfectly manageable through partial fraction decomposition.

The outcome is:

$$ \int_{0}^{\pi/2}\frac{d\theta}{\sin^3\theta+\cos^3\theta} = \color{red}{\frac{\pi}{3}+\frac{2\sqrt{2}}{3}\,\log\left(1+\sqrt{2}\right)}.$$

| cite | improve this answer | |
$\endgroup$
8
$\begingroup$

One may write $$ \begin{align} \int_0^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}&=\int_0^{\pi/2}\frac{\text{d}x}{(\cos x+\sin x)(\cos^2{x}-\cos x\sin x+\sin^2{x})} \\\\&=\frac{\sqrt{2}}2\int_0^{\pi/2}\frac{\text{d}x}{\cos(x-\frac{\pi}4)\:(1-\frac12\sin(2x))} \\\\&=\frac{\sqrt{2}}2\int_{-\pi/4}^{\pi/4}\frac{\text{d}u}{\cos u\:\left(1-\frac12\cos(2u)\right)} \\\\&=\sqrt{2}\int_0^{\pi/4}\frac{\cos u\:\text{d}u}{\left(1-\sin^2u\right)\:\left(\sin^2u+\frac12\right)} \\\\&=\sqrt{2}\int_0^{\sqrt{2}/2}\frac{\text{d}v}{\left(1-v^2\right)\:\left(v^2+\frac12\right)} \\\\&=\frac{\pi}3+\frac{2\sqrt{2}}{3}\log\left(1+\sqrt{2}\right). \end{align} $$

| 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.