Question-122506

Question Number 122506 by mathace last updated on 17/Nov/20

Answered by TANMAY PANACEA last updated on 17/Nov/20

\int \frac{l n (1 + {(x^{2} - 1)}^{2})}{x \sqrt{x^{2} - 1}} d x

x = s e c a

\int \frac{l n (1 + {t a n}^{4} a)}{s e c a t a n a} s e c a t a n a d a

I = \int_{0}^{\frac{π}{2}} l n (1 + {t a n}^{4} a) d a

I = \int_{0}^{\frac{π}{2}} l n (1 + \frac{1}{{t a n}^{4} a}) d a

I = \int_{0}^{\frac{π}{2}} l n (1 + {t a n}^{4} a) - 4 \int_{0}^{\frac{π}{2}} l n t a n a d a

I = I - 4 J \to J = 0

I = \int_{0}^{\frac{π}{2}} l n (1 + {t a n}^{4} a) d a

n o t r e a c h i n g t o g o a l

Commented by mathace last updated on 17/Nov/20

N o . I t h a s a f i n i t e v a l u e . C h e c k y o u r 4 t h l i n e .

Commented by TANMAY PANACEA last updated on 17/Nov/20

t h a n k y o u s i r

Commented by mindispower last updated on 17/Nov/20

w i t h e p l e a s u r

Commented by Dwaipayan Shikari last updated on 18/Nov/20

\int_{0}^{\frac{π}{2}} l o g (1 + {t a n}^{4} θ) d θ

= \int_{0}^{\frac{π}{2}} l o g ({s i n}^{4} θ + {c o s}^{4} θ) - 4 \int_{0}^{\frac{π}{2}} l o g (c o s θ) d θ

= \int_{0}^{\frac{π}{2}} l o g (1 - 2 {s i n}^{2} θ {c o s}^{2} θ) + 2 π l o g (2)

= \int_{0}^{\frac{π}{2}} l o g (1 - \frac{1}{2} {s i n}^{2} 2 θ) + 2 π l o g (2)

= \int_{0}^{\frac{π}{2}} l o g ({c o s}^{2} 2 θ + \frac{1}{2} {s i n}^{2} 2 θ) + 2 π l o g (2)

= π l o g (\frac{1 + \frac{1}{\sqrt{2}}}{2}) + π l o g (4)

= π l o g (2 + \sqrt{2})

Commented by mathace last updated on 18/Nov/20

W e l l d o n e, s i r .

Commented by Dwaipayan Shikari last updated on 18/Nov/20

\int_{0}^{\frac{π}{2}} l o g (a^{2} {s i n}^{2} θ + b^{2} {c o s}^{2} θ) = π l o g (\frac{a + b}{2})

I h a v e u s e d t h i s l e m m a . T h a n k i n g y o u

Commented by mnjuly1970 last updated on 18/Nov/20

t h a n k y o u

e x c e l l e n t . b r a v o \dots