Question

Evaluate:

`int_0^π(sin^4x + cos^4x)dx`

Answer 1

`int_0^π (sin^4x + cos^4x)dx`   ...`[∵ sin^4x + cos^4x = 1 - 1/2sin^2 2x = 1/4(3 + cos4x)]`

= `1/4int_0^π (3 + cos 4x)dx`

= `1/4[3x + (sin4x)/4]_0^π`

= `(3π)/4`