2

Evaluation of $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx$

$\bf{My\; Try::} $ Let $$I= \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{(\sin x+\cos x)(2-\sin 2x)}dx$$

So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{\sin\left(x+\frac{\pi}{4}\right)(2-\sin 2x)}dx$$

Put $\displaystyle x+\frac{\pi}{4} = t,$ Then $dx = dt$

So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t}{\sin t(2+\cos 2t)}dt=\frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-4+4}{\sin t(2+\cos 2t)}dt$$

So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-2}{\sin t}dt+\frac{2}{\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{1}{\sin t(2+\cos 2t)}dt$$

How can i solve it after that, Help required, Thanks

juantheron
  • 53,015

2 Answers2

2

HINT:

If $f(x)=\sin^2x\cos^2x=\dfrac{\sin^22x}4,f\left(\dfrac\pi4+0-x\right)=\dfrac{\cos^22x}4=\dfrac{(1-2\sin^2x)^2}4$

If $g(x)=\sin^3x+\cos^3x,g\left(\dfrac\pi4+0-x\right)=\dfrac{2\cos x(\cos^2x+3\sin^2x)}{2\sqrt2}=\dfrac1{\sqrt2}\cdot\dfrac{(1-\sin^2x)(1+2 \sin^2x)}{\cos x}$

Now use $$\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$ and choose $\sin x=u$

lab bhattacharjee
  • 274,582
0

Hint

From where you left off:

The first integrand is the same as $$-\csc t-4\sin t+4\sin^3t$$

The second integral will be evaluated using the substitution $u=\cos t$ followed by a dose of partial fractions.

David Quinn
  • 34,121