Definite Integral $\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{8\cos2x}{(\tan x+\cot x)^3}\,dx$.

Question

$$\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\dfrac{8\cos2x}{(\tan x+\cot x)^3}\,dx$$

The problem is attached as a photograph!

so I started this, by taking cot(x) as 1/tan(x) and then after simplifying, I got, 8{2sin³(x)cos^5(x)- sin³(x)cos³(x)}

After converting these to Sin(2x) I got,

2cos²x(sin2x)³ - (sin2x)³

Taking sin³2x common, we get,

Sin³2x(cos2x)

Taking sin2x = t Hence (2cos2x)dx=dt

I get (1/2)* integral of (t³)dt

Which gives me t⁴/8 as the ans

Now substitution part. It's no where near to the options. MI going wrong somewhere? Or the whole method itself is not correct/cannot be put to use here?

I have just started studying integration. So pls help me out. PS: sorry for making this all so boring!

Answer 1

$\tan x+\cot x=\dfrac{2}{\sin2x}$ so $$I=\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} 8\cos2x\sin^32x\,dx=\dfrac18\sin^42x\Big|_{\frac{\pi}{12}}^{\frac{\pi}{4}} = \color{blue}{\dfrac{15}{128}}$$

Answer 2

$$\frac{8\cos2x}{(\tan{x}+\cot{x})^3}=8\cos2x\sin^3x\cos^3x=\cos2x\sin^32x=$$ $$=\frac{\cos2x(3\sin2x-\sin6x)}{4}=\frac{1}{8}(2\sin4x-\sin8x).$$ Also, we have $$\cos2x\sin^32xdx=\frac{1}{2}\sin^32xd(\sin2x).$$ I hope now it's clear.