Question

$\int$ sin⁴x dx, = ax + b sin 2x + c sin 4x + d then 8a + 4b + 32 c =

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\int \sin ^ { 4 } x d x$ = $\int \left( \frac { 1 - \cos 2 x } { 2 } \right) ^ { 2 } \mathrm { dx }$

= $\frac { 1 } { 4 } \int \left( 1 + \cos ^ { 2 } 2 x - 2 \cos 2 x \right) d x$

=

= $\frac { x } { 4 } - \frac { \sin 2 x } { 4 } + \frac { x } { 8 } + \frac { \sin 4 x } { 32 } + C$= $\frac { 3 x } { 8 } - \frac { \sin 2 x } { 4 } + \frac { \sin 4 x } { 32 } + C$