Question-44573

Question Number 44573 by Raj Singh last updated on 01/Oct/18

Commented by maxmathsup by imad last updated on 01/Oct/18

l e t A = \int_{0}^{\frac{π}{4}} l n (s i n (2 θ)) d θ \Rightarrow A =_{2 θ = t} \int_{0}^{\frac{π}{2}} l n (s i n (t)) \frac{d t}{2}

= \frac{1}{2} \int_{0}^{\frac{π}{2}} l n (s i n t) d t l e t I = \int_{0}^{\frac{π}{2}} l n (s i n (t)) d t a n d J = \int_{0}^{\frac{π}{2}} l n (c o s t) d t

c h a n g e m e n t t = \frac{π}{2} - u s h o w t h a t I = J \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (l n (s i n t) + l n (c o s t)) d t

= \int_{0}^{\frac{π}{2}} l n (\frac{1}{2} s i n (2 t)) d t = - \frac{π}{2} l n (2) + \int_{0}^{\frac{π}{2}} l n (s i n (2 t)) d t b u t

\int_{0}^{\frac{π}{2}} l n (s i n (2 t) d t =_{2 t = u} \int_{0}^{π} l n (s i n (u)) \frac{d u}{2}

= \frac{1}{2} {\int_{0}^{\frac{π}{2}} l n (s i n (u)) d u + \int_{\frac{π}{2}}^{π} l n (s i n u) d u} = \frac{1}{2} I + \frac{1}{2} \int_{\frac{π}{2}}^{π} l n (s i n u) d u b u t

\int_{\frac{π}{2}}^{π} l n (s i n (u)) d u =_{u = \frac{π}{2} + θ} \int_{0}^{\frac{π}{2}} l n (c o s θ) d θ = I \Rightarrow

2 I = - \frac{π}{2} l n (2) + I \Rightarrow I = - \frac{π}{2} l n (2) \Rightarrow A = - \frac{π}{4} l n (2) .

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Oct/18

I = \int_{0}^{\frac{π}{4}} l n s i n 2 θ d θ

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

= \int_{0}^{\frac{π}{2}} l n c o s 2 θ d θ

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

2 I = \int_{0}^{\frac{π}{4}} l n (\frac{s i n 4 θ}{2}) d θ

2 I = \int_{0}^{\frac{π}{4}} l n s i n 4 θ d θ - l n 2 \int_{0}^{\frac{π}{4}} d θ

2 θ = t s o d θ = \frac{d t}{2}

2 I = \frac{1}{2} \int_{0}^{\frac{π}{2}} l n s i n 2 t d t - l n 2 \times \frac{π}{4}

2 I = \frac{1}{2} \times 2 \int_{0}^{\frac{π}{4}} l n s i n 2 t d t - l n 2 \times \frac{π}{4}

2 I = I - l n 2 \times \frac{π}{4}

I = - l n 2 \times \frac{π}{4}