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}

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}

Ashley Bell 2022-01-01 Answered

I want to solve these Integrals
1.

\int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{\sqrt{2}} x} d x

2.

\int_{0}^{\frac{π}{2}} \frac{1}{{(\sqrt{2} \cos^{2} x + \sin^{2} x)}^{2}}

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Expert Answer

macalpinee3

Answered 2022-01-02 Author has 29 answers

For the second integral:
Substitute $t = \tan x$ in order to get:
$\int_{0}^{\frac{π}{2}} \frac{1}{{(\sqrt{2} \cos^{2} x + \sin^{2} x)}^{2}} d x = \int_{0}^{\infty} \frac{t^{2} + 1}{{(\sqrt{2} + t^{2})}^{2}}$
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 easy to calculate that $a = \frac{\sqrt{2} + 1}{2 \sqrt{2}}$ and $b = \frac{\sqrt{2} - 1}{2 \sqrt{2}}$ , from the system of equations
$a + b = 1$
$a - b = \frac{1}{\sqrt{2}}$
Using the previous, we get
$\int_{0}^{\infty} \frac{t^{2} + 1}{{(\sqrt{2} + t^{2})}^{2}} d t = a \int_{0}^{\infty} \frac{1}{\sqrt{2} + t^{2}} d t + b \int_{0}^{\infty} \frac{t^{2} - \sqrt{2}}{{(\sqrt{2} + t^{2})}^{2}}$
$\int_{0}^{\infty} \frac{t^{2} + 1}{(\sqrt{2} + t^{2})^{2}} d t = a \frac{\arctan \frac{t}{\sqrt[4]{2}}}{\sqrt[4]{2}} |_{0}^{\infty} - b \frac{t}{t^{2} + \sqrt{2}} |_{0}^{\infty}$
$\int_{0}^{\infty} \frac{t^{2} + 1}{(\sqrt{2} + t^{2})^{2}} d t = a \frac{\frac{π}{2}}{\sqrt[4]{2}}$
$\int_{0}^{\infty} \frac{t^{2} + 1}{(\sqrt{2} + t^{2}} d t = \frac{\sqrt{2} + 1}{4 \sqrt{2}} \frac{π}{\sqrt[4]{2}}$

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Timothy Wolff

Answered 2022-01-03 Author has 26 answers

Let

I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{\sqrt{2}} x} d x

Using

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

So

I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \cot^{\sqrt{2}} x} d x = \int_{0}^{\frac{π}{2}} \frac{\tan^{\sqrt{2}} x}{1 + \tan^{\sqrt{2}} x} d x

Now Adding
So we get

2 I = \int_{0}^{\frac{π}{2}} 1 \cdot d x = \frac{π}{2} \Rightarrow I = \frac{π}{4}

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karton

Answered 2022-01-09 Author has 368 answers

Set
$\begin{matrix} I (a, b, n) = \int_{0}^{\frac{π}{2}} \frac{1}{(a \cos^{2} x + b \sin^{2} x)^{n}} d x \\ \frac{d I (a, b, n)}{d a} = - n \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{(a \cos^{2} x + b \sin^{2} x)^{n + 1}} \\ \frac{d I (a, b, n)}{d b} = - n \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{(a \cos^{2} x + b \sin^{2} x)^{n + 1}} \\ Therefore \\ \frac{d I (a, b, n)}{d a} + \frac{d I (a, b, n)}{d b} = - n I (a, b, n + 1) \\ we have \\ I (a, b, n + 1) = - \frac{1}{n} (\frac{d I (a, b, n)}{d a} + \frac{d I (a, b, n)}{d b}) \\ Now we should compute I (a, b, 1) \\ I (a, b, 1) = \int_{0}^{\frac{π}{2}} \frac{1}{a \cos^{2} x + b \sin^{2} x} d x = \int_{0}^{\frac{/ p i}{2}} \frac{1 + \tan^{2} x}{a + b \tan^{2} x} d x \\ Set u = \tan x, thus \\ I (a, b, 1) = \int_{0}^{\infty} \frac{1}{a + b u^{2}} d x = \frac{π}{2 \sqrt{a b}} \\ so \\ \frac{d I (a, b, 1)}{d a} = \frac{π \sqrt{b}}{4 a \sqrt{a} b} \\ similarly \\ \frac{d I (a, b, 1)}{d b} = \frac{π \sqrt{a}}{4 b \sqrt{b} a} \\ apply \\ I (a, b, 2) = - (\frac{d I (a, b, 1)}{d a} + \frac{d I (a, b, 1)}{d b}) = \frac{π (a + b)}{4 a b \sqrt{a b}} \\ set a = \sqrt{2} and b = 1, finally \\ \int_{0}^{\frac{π}{2}} \frac{1}{(\sqrt{2} \cos^{2} x + \sin^{2} x)^{2}} = \frac{π (1 + \sqrt{2})}{4 \sqrt[4]{8}} \end{matrix}$