Question

$\int _ { 0 } ^ { \pi / 4 } \frac { 4 \sin 2 \theta d \theta } { \sin ^ { 4 } \theta + \cos ^ { 4 } \theta } =$

• A. $\pi / 4$
• B. $\pi / 2$
• C. $\pi$
• D. None of these

Answer 1

Answer: (C)

$\pi$

Answer 2

$4 \int _ { 0 } ^ { \pi / 4 } \frac { \sin 2 \theta d \theta } { \sin ^ { 4 } \theta + \cos ^ { 4 } \theta } = 4 \int _ { 0 } ^ { \pi / 4 } \frac { 2 \sin \theta \cos \theta d \theta } { \sin ^ { 4 } \theta + \cos ^ { 4 } \theta }$

$= 4 \int _ { 0 } ^ { \pi / 4 } \frac { 2 \tan \theta \sec ^ { 2 } \theta d \theta } { \tan ^ { 4 } \theta + 1 }$

{Dividing numerator and denominator by $\cos ^ { 4 } \theta$}

Now put$4 \int _ { 0 } ^ { 1 } \frac { d t } { t ^ { 2 } + 1 } = 4 \left[ \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 } = 4 \left[ \frac { 1 } { 4 } \pi - 0 \right] = \pi$.