calculate ∫_0 ^(π/4) (dt/((1+sin^2 t)^2 )) .


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Question Number 32354 by abdo imad last updated on 23/Mar/18

$c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + {s i n}^{2} t)}^{2}} .$
Commented by abdo imad last updated on 28/Mar/18

$l e t p u t I = \int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + {s i n}^{2} t)}^{2}}$ $I = \int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + \frac{1 - c o s (2 t)}{2})}^{2}} = 4 \int_{0}^{\frac{π}{4}} \frac{d t}{{(3 - c o s (2 t))}^{2}}$ $=_{2 t = x} 4 \int_{0}^{\frac{π}{2}} \frac{1}{{(3 - c o s x)}^{2}} \frac{d x}{2} = 2 \int_{0}^{\frac{π}{2}} \frac{d x}{{(3 - c o s x)}^{2}}$ $c h . t a n (\frac{x}{2}) = t g i v e$ $I = 2 \int_{0}^{1} \frac{1}{{(3 - \frac{1 - t^{2}}{1 + t^{2}})}^{2}} \frac{2 d t}{1 + t^{2}} = 4 \int_{0}^{1} \frac{d t}{(1 + t^{2}) \frac{{(3 + 3 t^{2} - 1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}}}$ $= 4 \int_{0}^{1} \frac{1 + t^{2}}{{(2 + 4 t^{2})}^{2}} d t = \int_{0}^{1} \frac{1 + t^{2}}{{(1 + 2 t^{2})}^{2}} d t$ $= \int_{0}^{1} \frac{1 + 2 t^{2} - t^{2}}{{(1 + 2 t^{2})}^{2}} d t = \int_{0}^{1} \frac{d t}{(1 + 2 t^{2})} d t - \int_{0}^{1} \frac{t^{2}}{{(1 + 2 t^{2})}^{2}} d t$ $\int_{0}^{1} \frac{d t}{1 + 2 t^{2}} d t =_{t \sqrt{2} = u} \int_{0}^{\sqrt{2}} \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{2}} = \frac{1}{\sqrt{2}} a r t a n (\sqrt{2})$ $\int_{0}^{1} \frac{t^{2}}{{(1 + 2 t^{2})}^{2}} d t =_{t \sqrt{2} = u} \int_{0}^{\sqrt{2}} \frac{1}{{(1 + u^{2})}^{2}} \frac{u^{2}}{2} \frac{d u}{\sqrt{2}}$ $= \frac{1}{2 \sqrt{2}} \int_{0}^{\sqrt{2}} \frac{u^{2}}{{(1 + u^{2})}^{2}} d u b u t$ $\int_{0}^{\sqrt{2}} \frac{u^{2}}{{(1 + u^{2})}^{2}} d u = - \frac{1}{2} \int_{0}^{\sqrt{2}} u . (\frac{- 2 u}{{(1 + u^{2})}^{2}}) d u b y p a r t s$ $= - \frac{1}{2} ({[\frac{u}{1 + u^{2}}]}_{0}^{\sqrt{2}} - \int_{0}^{\sqrt{2}} \frac{u}{1 + u^{2}} d u)$ $= - \frac{1}{2} \frac{\sqrt{2}}{3} + \frac{1}{4} \int_{0}^{\sqrt{2}} \frac{2 u}{1 + u^{2}} d u = \frac{- \sqrt{2}}{6} + \frac{1}{4} {[l n (1 + u^{2})]}_{0}^{\sqrt{2}}$ $= \frac{- \sqrt{2}}{6} + \frac{1}{4} l n (3) \Rightarrow \int_{0}^{1} \frac{t^{2}}{{(1 + 2 t^{2})}^{2}} d t = \frac{1}{2 \sqrt{2}} (\frac{- \sqrt{2}}{6} + \frac{1}{4} l n 3)$ $= \frac{- 1}{12} + \frac{l n 3}{8 \sqrt{2}}$ $I = \frac{1}{\sqrt{2}} a r c t a n (\sqrt{2}) + \frac{1}{12} - \frac{l n 3}{8 \sqrt{2}} .$
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