If $$I=\int^{\pi/4}_{-\pi/4}\frac{\sqrt{\cos x}}{\left(\sqrt{\cos x}+\sqrt{\sin x}\right)^5}\,dx$$, then what is $27I^2$?
Plan:
$$I=\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\sec^2 x}{\left(1+\sqrt{\tan x}\right)^5}\,dx$$
Put $\tan x=t$ and $\sec^2 dx=dt$
$$I=\int^{1}_{-1}\frac{2t}{(1+t)^5}\,dt$$
How do i solve it help me
Let me bother about the last step of yours: $$ I=\int_{-1}^{1} \frac{2t dt}{(1+t)^5}.$$ Sorry, this is Improper and divergent (infinite).
One way, an Integral $\int_{a}^{b} f(x) dx$ becomes Imporper if it diverges at either of the limits $x=a,b$ or in the domain $(a,b)$. Interestingly, if $ f(x) \sim (x-a)^{-k}$, $I$ could be convergent (finite) if $0< k <1$, otherwise it will diverge and become infinite. Look at the following example of Improper but convergent integrals. $$\int_{0}^{1} \frac{dx}{x^{0.99}}, \int_{0}^{1} \frac{dx}{\sqrt{1-x}},~~ \int_{-1}^{1} \frac{dx}{\sqrt{1-x^2}}.$$ Next, the following improper integral $$\int_{0}^{1} \frac{dx}{x^{1.01}}, \int_{0}^{1} \frac{dx}{(1-x)^{3/2}},~~ \int_{-1}^{1} \frac{dx}{(1-x^2)^{3/2}}.$$ are diverget (infinite). Also mind others comments.
In your original question if you make the the domain of integration as $[0,\pi/4]$, and write $\tan x= t^2$ your integral will be real and finite: $$ J=\int_{0}^{1} \frac{2t dt}{(1+t)^5} = \int_{1}^{2} \frac{2(u-1)}{u^5 }du=\frac{11}{96}.$$
Am i correct ?
– learningstudent May 27 '19 at 17:43