Let, n ≥ 2 be a natural number and 0 < θ < (π)/(2) . Then ∫ (si n ^...

$∫ \dfrac{\left(\sin^{n}\theta - sin\theta\right)\dfrac{1}{n}cos\theta}{\sin^{n + 1}\theta}d\theta$

Put, $sin\theta = t ⇒ cos\thetad\theta = dt$

$=$\begin{cases}∫ \dfrac{\left({t}^{n} - t\right)\dfrac{1}{n}dt}{{t}^{n + 1}} & \end{cases}$$<br /><br />$= ∫ \dfrac{t\left(1 - \dfrac{1}{{t}^{n - 1}}\right)\dfrac{1}{n}}{{t}^{n + 1}}dt$<br /><br />$= ∫ \dfrac{{\left(1 - \dfrac{1}{{t}^{n - 1}}\right)}^{\dfrac{1}{n}}}{{t}^{n}}dt$<br /><br />Put$1 - \dfrac{1}{{t}^{n - 1}} = z$</p><p>$⇒ \dfrac{\left(n - 1\right)}{{t}^{n}}dt = dz$<br /><br />$⇒ I = \dfrac{1}{n - 1} ∫ z\dfrac{1}{n}dz$</p><p>Using$∫ {x}^{n}dx = \dfrac{{x}^{n + 1}}{n + 1} + c$</p><p>$⇒ I = \dfrac{{z}^{\dfrac{1}{n} + 1}}{\left(\dfrac{1}{n} + 1\right)\left(n - 1\right)} + c$</p><p>$⇒ I = \dfrac{n{\left(1 - {t}^{1 - n}\right)}^{\dfrac{1}{n} + 1}}{{n}^{2} - 1} + c,$where$c$is the constant of integration.<br />$⇒ I = \dfrac{n}{{n}^{2} - 1}\left(1 - \dfrac{1}{\sin^{n - 1}\theta}\right)\dfrac{n + 1}{n} + c .$