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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\int_0^{\pi /8} {{{\cos }^3}4\theta d\theta } = $
  • A
    $\frac{2}{3}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{1}{3}$
  • ✓
    $\frac{1}{6}$
✓

Answer

Correct option: D.
$\frac{1}{6}$
d
(d) Let $I = \int_0^{\pi /8} {{{\cos }^3}4\theta \,d\theta = \int_0^{\pi /8} {\,{{\cos }^2}4\theta .\cos 4\theta \,d\theta } } $

$I = \int_0^{\pi /8} {\,(1 - {{\sin }^2}4\theta )\cos 4\theta \,d\theta } $

Put $\sin 4\theta = t \Rightarrow \cos 4\theta \,d\theta = \frac{{dt}}{4}$

When $\theta = 0 \to \frac{\pi }{8},$ then $t = 0 \to 1$

$\therefore$  $I = \frac{1}{4}\int_0^1 {(1 - {t^2})dt = \frac{1}{4}} \left[ {t - \frac{{{t^3}}}{3}} \right]_0^1 = \frac{1}{6}$.

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