(C) Let $I=\int_0^{\pi / 4} \sec ^7 \theta \cdot \sin ^3 \theta d \theta=\int_0^{\pi / 4} \tan ^3 \theta \sec ^4 \theta d \theta$ Put $\tan \theta= t \Rightarrow \sec ^2 \theta d \theta= dt$ When $\theta=0, t=0$ and when $\theta=\frac{\pi}{4}, t=1$ $\therefore \quad I=\int_0^1 t^3\left(1+t^2\right) d t=\left[\frac{t^4}{4}+\frac{t^6}{6}\right]_0^1=\frac{5}{12}$
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