Question

If the value of the integral $\displaystyle \int _{0}^{\frac{\pi }{2}} c o t\frac{\theta }{2}\left(1 - \left(c o s\right)^{4} \theta \right)d\theta $ is equal to $\lambda ,$ then the value of $3\lambda $ is equal to

Answer 1

Let $I=\displaystyle \int _{0}^{\frac{\pi }{2}}cot\frac{\theta }{2}\left(1 - c o s \theta \right)\left(1 + c o s \theta + c o s^{2} \theta + c o s^{3} \theta \right)d\theta $
So, $I=\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{c o s \frac{\theta }{2}}{s i n \frac{\theta }{2}}\cdot 2sin^{2}\frac{\theta }{2}\left(1 + c o s \theta + \left(cos\right)^{2} \theta + \left(cos\right)^{3} ⁡ \theta \right)d\theta $
$=\displaystyle \int _{0}^{\frac{\pi }{2}}sin \theta \left(1 + cos \theta + \left(cos\right)^{2} \theta + \left(cos\right)^{3} \theta \right)d\theta $
Let $cos \theta = t$
$\Rightarrow I=\displaystyle \int _{0}^{1}\left(1 + t + t^{2} + t^{3}\right)dt=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}$
Hence, $\lambda =\frac{25}{12}\Rightarrow 3\lambda =\frac{25}{4}=6.25$