pi-3-pi-3-x-sin-x-cos-2-x-dx-

Question Number 55772 by gunawan last updated on 04/Mar/19

\underset{- π / 3}{\int^{π / 3}} \frac{x \sin x}{\cos^{2} x} d x =

Commented by maxmathsup by imad last updated on 04/Mar/19

l e t A = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{x s i n x}{{c o s}^{2} x} d x \Rightarrow A = 2 \int_{0}^{\frac{π}{3}} \frac{x s i n x}{{c o s}^{2} x} d x b y p a r t s u = x a n d v^{^{'}} = \frac{s i n x}{{c o s}^{2} x}

A = 2 {{[\frac{x}{c o s x}]}_{0}^{\frac{π}{3}} - \int_{0}^{\frac{π}{3}} \frac{d x}{c o s x}} = 2 {\frac{π}{3 \frac{1}{2}} - \int_{0}^{\frac{π}{3}} \frac{d x}{c o s x}}

= \frac{4 π}{3} - 2 \int_{0}^{\frac{π}{3}} \frac{d x}{c o s x} b u t \int_{0}^{\frac{π}{3}} \frac{d x}{c o s x} =_{t a n (\frac{x}{2}) = t} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{\frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\frac{1}{\sqrt{3}}} \frac{2 d t}{1 - t^{2}} = \int_{0}^{\frac{1}{\sqrt{3}}} {\frac{1}{1 - t} + \frac{1}{1 + t}} d t = {[l n ∣ \frac{1 + t}{1 - t} ∣]}_{0}^{\frac{1}{\sqrt{3}}} = l n ∣ \frac{\sqrt{3} + 1}{\sqrt{3} - 1} ∣ \Rightarrow

A = \frac{4 π}{3} - 2 l n (\frac{\sqrt{3} + 1}{\sqrt{3} - 1}) .

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19

\int x t a n x s e x d x

x \int t a n x s e c x d x - \int [\frac{d x}{d x} \int t a n x s e c x d x] d x

x s e c x - \int s e c x d x

x s e c x - l n (s e c x + t a n x)

∣ x s e c x - l n (s e c x + t a n x) ∣_{\frac{- π}{3}}^{\frac{π}{3}}

[{\frac{π}{3} s e c \frac{π}{3} - l n (s e c \frac{π}{3} + t a n \frac{π}{3})} - {\frac{- π}{3} s e c (\frac{- π}{3}) - l n (s e c \frac{- π}{3} + t a n \frac{- π}{3})}]

= [{\frac{π}{3} \times 2 - l n (2 + \sqrt{3})} - {\frac{- 2 π}{3} - l n (2 - \sqrt{3})}]

= \frac{4 π}{3} + l n (\frac{2 - \sqrt{3}}{2 + \sqrt{3}})

p l s c h e c k

Commented by gunawan last updated on 04/Mar/19

Answer is true Sir

Thanks

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19

t h a n k y o u s i r t o m a k e o u r b r a i n a c t i v e a n d a g i l e \dots