Two complex definite Integrals

Question

I want to solve these Integrals

1.

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan^{\sqrt {2}}x} dx$$

2. $$\int_{0}^{\frac{\pi}{2}}\frac{1}{(\sqrt{2}\cos^2x+\sin^2x)^2}$$ Thanks

Answer 1

For the second integral:

Substitute $t=\tan{x}$ in order to get: $$\int_{0}^{\frac{\pi}{2}}\frac{1}{(\sqrt{2}\cos^2x+\sin^2x)^2} dx=\int_{0}^{\infty}\frac{t^2+1}{(\sqrt{2}+t^2)^2}dt.$$ Observe that $$d(-\frac{t}{t^2+\sqrt{2}})=\frac{t^2-\sqrt{2}}{(t^2+\sqrt{2})^2}.$$ Now write $t^2+1$ as the following sum: $$a(t^2+\sqrt{2})+b(t^2-\sqrt{2}).$$ It is easy to calculate that $\boxed{a=\frac{\sqrt{2}+1}{2\sqrt{2}}}$ and $\boxed{b=\frac{\sqrt{2}-1}{2\sqrt{2}}}$, from the system of equations $$a+b=1,$$ $$a-b=\frac{1}{\sqrt{2}}.$$ Using the previous, we get $$\int_{0}^{\infty}\frac{t^2+1}{(\sqrt{2}+t^2)^2}dt=a\int_{0}^{\infty}\frac{1}{\sqrt{2}+t^2}dt+b\int_{0}^{\infty}\frac{t^2-\sqrt{2}}{(\sqrt{2}+t^2)^2}dt,$$ $$\int_{0}^{\infty}\frac{t^2+1}{(\sqrt{2}+t^2)^2}dt=a\frac{\arctan{\frac{t}{\sqrt[4]{2}}}}{\sqrt[4]{2}}|_0^\infty-b\frac{t}{t^2+\sqrt{2}}|_0^\infty,$$ $$\int_{0}^{\infty}\frac{t^2+1}{(\sqrt{2}+t^2)^2}dt=a\frac{\frac{\pi}{2}}{\sqrt[4]{2}},$$ $$\boxed{\int_{0}^{\infty}\frac{t^2+1}{(\sqrt{2}+t^2)^2}dt=\frac{\sqrt{2}+1}{4\sqrt{2}}\frac{\pi}{\sqrt[4]{2}}.}$$

Answer 2

Let $$I = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan^{\sqrt{2}}x}dx.....(1)$$

Using $$\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$$

So $$I = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot^{\sqrt{2}}x}dx = \int_{0}^{\frac{\pi}{2}}\frac{\tan ^{\sqrt{2}}x}{1+\tan^{\sqrt{2}}x}dx....(2)$$

Now Adding $(1)$ and $(2)$

So we get $$2I = \int_{0}^{\frac{\pi}{2}}1\cdot dx = \frac{\pi}{2}\Rightarrow I = \frac{\pi}{4}$$

Answer 3

Set $$I(a,b,n)=\int_{0}^{\frac{\pi}{2}}\frac{1}{(a\cos^2x+b\sin^2x)^n}dx$$ $$\frac{\partial I(a,b,n)}{\partial a}=-n\int_{0}^{\frac{\pi}{2}}\frac{\cos^2x}{(a\cos^2x+b\sin^2x)^{n+1}}dx$$ $$\frac{\partial I(a,b,n)}{\partial b}=-n\int_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{(a\cos^2x+b\sin^2x)^{n+1}}dx$$ Therefore $$\frac{\partial I(a,b,n)}{\partial a}+\frac{\partial I(a,b,n)}{\partial b}=-nI(a,b,n+1)$$ we have

$$I(a,b,n+1)=-\frac{1}{n}\left(\frac{\partial I(a,b,n)}{\partial a}+\frac{\partial I(a,b,n)}{\partial b}\right)\tag 1$$ Now we should compute $I(a,b,1)$ $$I(a,b,1)=\int_{0}^{\frac{\pi}{2}}\frac{1}{a\cos^2x+b\sin^2x}dx=\int_{0}^{\frac{\pi}{2}}\frac{1+\tan^2x}{a+b\tan^2x}dx$$ Set $u= \tan x$, thus $$I(a,b,1)=\int_{0}^{\infty}\frac{1}{a+bu^2}dx=\frac{\pi}{2\sqrt{ab}}$$ so $$\frac{\partial I(a,b,1)}{\partial a}=\frac{\pi\sqrt{b}}{4a\sqrt{a}\,b}$$ similarly $$\frac{\partial I(a,b,1)}{\partial b}=\frac{\pi\sqrt{a}}{4b\sqrt{b}\,a}$$ apply $(1)$ $$I(a,b,2)=-\left(\frac{\partial I(a,b,1)}{\partial a}+\frac{\partial I(a,b,1)}{\partial b}\right)=\frac{\pi(a+b)}{4ab\sqrt{ab}}$$ set $a=\sqrt{2}$ and $b=1$, finally

$$\int_{0}^{\frac{\pi}{2}}\frac{dx}{(\sqrt{2}\cos^2x+\sin^2x)^2}=\frac{\pi(1+\sqrt{2})}{4\sqrt[4]{8}}$$