MCQ
$\int_0^{\pi /4} {{{\sec }^7}\theta {{\sin }^3}\theta } \,d\theta = $
- A$\frac{1}{{12}}$
- B$\frac{3}{{12}}$
- ✓$\frac{5}{{12}}$
- Dએકપણ નહીં.
$=\int_0^{\pi /4} {\frac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}.{{\sec }^4}\theta \,d\theta } $
Putting $\tan \theta = t,$ it reduces to
$\int_0^1 {{t^3}(1 + {t^2})\,dt} =$$ \left| {\frac{{{t^4}}}{4} + \frac{{{t^6}}}{6}} \right|_0^1 = \frac{5}{{12}}$.
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