QUESTION

# $\int\limits_0^{\dfrac{\pi }{8}} {{{\cos }^3}4\theta } d\theta$ is equal to(A) $\dfrac{5}{3}$ (B) $\dfrac{5}{4}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{6}$

Hint: Here we have to solve the integral by substituting appropriately and then change the limits as per the variable.

Let $I = \int\limits_0^{\dfrac{\pi }{8}} {{{\cos }^3}4\theta } d\theta$
So this can be written as,
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{8}} {{{\cos }^2}4\theta \cdot } \cos 4\theta d\theta$
Now you know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta$
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{8}} {\left( {1 - {{\sin }^2}4\theta } \right)} \cdot \cos 4\theta d\theta$
Now let $\sin 4\theta = t$ $\to$ (1)
Limits:
Lower limit; when $\theta = 0 \Rightarrow t = \sin(0) = 0$
Upper limit; when $\theta = \dfrac{\pi }{8} \Rightarrow t = \sin \dfrac{\pi }{2} = 1$

Now differentiate equation (1) w.r.t $\theta$
$4\cos 4\theta d\theta = dt$
Substituting these, we get,
$\Rightarrow I = \int\limits_0^1 {\left( {1 - {t^2}} \right) \cdot \dfrac{{dt}}{4}}$
Apply the integration
$\Rightarrow I = \dfrac{1}{4}\left[ {t - \dfrac{{{t^3}}}{3}} \right]_0^1$
Apply the integration limits
$\Rightarrow I = \dfrac{1}{4}\left[ {1 - \dfrac{{{1^3}}}{3} - 0 + 0} \right]_0^1$
$\Rightarrow I = \dfrac{1}{4} \times \dfrac{2}{3} = \dfrac{1}{6}$
So option (D) is the correct answer.

Note: In these types of problems, when a part of an existing integrad is equated to another variable, the limits also change accordingly but one usually forgets that part and substitutes the original limits which will lead to a wrong answer most of the time.