calculate-0-arctan-x-2-1-x-2-dx-

Question Number 59050 by maxmathsup by imad last updated on 04/May/19

c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (x^{2})}{1 + x^{2}} d x

Answered by tanmay last updated on 04/May/19

x^{2} = t a n a

2 x d x = {s e c}^{2} a d a

d x = \frac{{s e c}^{2} a d a}{2 \sqrt{t a n a}} d a

\int_{0}^{\frac{π}{2}} \frac{a}{1 + t a n a} \times \frac{{s e c}^{2} a}{2 \sqrt{t a n a}} d a

\int_{0}^{\frac{π}{2}} \frac{a}{{c o s}^{2} a (} \times \frac{c o s a}{s i n a + c o s a}) \times \frac{\sqrt{c o s a}}{2 \sqrt{s i n a}} d a

\int_{0}^{\frac{π}{2}} \frac{a d a}{2 \sqrt{s i n a c o s a}} \times \frac{1}{(s i n a + c o s a)}

\int_{0}^{\frac{π}{2}} \frac{a d a}{\sqrt{s i n 2 a} \times (s i n a + c o s a)} = I

I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - a)}{\sqrt{s i n 2 (\frac{π}{2} - a)} \times (s i n (\frac{π}{2} - a) + c o s (\frac{π}{2} - a)} d a

I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - a)}{\sqrt{s i n 2 a} \times (c o s a + s i n a)} d a

2 I = \int_{0}^{\frac{π}{2}} \frac{\frac{π}{2}}{\sqrt{s i n 2 a} \times (s i n a + c o s a)}

\frac{4 I}{π} = \int_{0}^{\frac{π}{2}} \frac{d a}{\sqrt{s i n 2 a} \times \sqrt{1 + s i n 2 a}}

t = t a n a \to d t = {s e c}^{2} a d a \to d a = \frac{d t}{1 + t^{2}}

\frac{4 I}{π} = \int_{0}^{\infty} \frac{1}{\sqrt{\frac{2 t}{1 + t^{2}}}} \times \frac{1}{\sqrt{1 + \frac{2 t}{1 + t^{2}}}} \times \frac{d t}{1 + t^{2}}

\frac{4 I}{π} = \int_{0}^{\infty} \frac{d t}{\sqrt{2 t} \times (1 + t)}

\frac{4 \sqrt{2} I}{π} = \int_{0}^{\infty} \frac{d t}{\sqrt{t} (1 + t)}

k^{2} = t \to d t = 2 k d k

\frac{4 \sqrt{2} I}{π} = \int_{0}^{\infty} \frac{2 k d k}{k (1 + k^{2})}

\frac{2 \sqrt{2}}{π} \times I = \int_{0}^{\infty} \frac{d k}{1 + k^{2}}

\frac{2 \sqrt{2}}{π} \times I =∣ {t a n}^{- 1} k ∣_{0}^{\infty}

I = \frac{π}{2 \sqrt{2}} \times \frac{π}{2} \to I = \frac{π^{2}}{4 \sqrt{2}}

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