find-I-n-0-pi-2-1-cos-nx-sin-2-nx-dx-

Question Number 60685 by maxmathsup by imad last updated on 24/May/19

f i n d I_{n} = \int_{0}^{\frac{π}{2}} \frac{1 - c o s (n x)}{{s i n}^{2} (n x)} d x

Commented by maxmathsup by imad last updated on 25/May/19

w e h a v e I_{n} = \int_{0}^{\frac{π}{2}} \frac{2 {s i n}^{2} (\frac{n x}{2})}{4 {s i n}^{2} (\frac{n x}{2}) {c o s}^{2} (\frac{n x}{2})} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{d x}{{c o s}^{2} (\frac{n x}{2})}

=_{\frac{n x}{2} = t} \frac{1}{2} \int_{0}^{\frac{n π}{4}} \frac{1}{{c o s}^{2} t} \frac{2}{n} d t = \frac{1}{n} \int_{0}^{\frac{n π}{4}} \frac{d t}{\frac{1 + c o s (2 t)}{2}} = \frac{2}{n} \int_{0}^{\frac{n π}{4}} \frac{d t}{1 + c o s (2 t)}

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

= \frac{1}{n} \int_{0}^{t a n (\frac{n π}{4})} d u = \frac{1}{n} t a n (\frac{n π}{4})

★ I_{n} = \frac{1}{n} t a n (n \frac{π}{4}) ★ w i t h n > 0 .

Answered by tanmay last updated on 24/May/19

\int_{0}^{\frac{π}{2}} \frac{2 {s i n}^{2} \frac{n x}{2}}{{(2 s i n \frac{n x}{2} c o s \frac{n x}{2})}^{2}} d x

\frac{1}{2} \int_{0}^{\frac{π}{2}} {s e c}^{2} \frac{n x}{2} d x

\frac{1}{2} \times ∣ t a n \frac{n x}{2} ∣_{0}^{\frac{π}{2}} \times \frac{2}{n}

\frac{1}{n} \times t a n \frac{n π}{4}

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