Question

请问这两道积分的题目怎么做

请问这两道积分的题目怎么做方便的话请给我详细点的解答，谢谢(*°∀°)=3

Answer 1

cos⁴x=(cos²x)²=[(1+cos2x)/2]²
=(1/4)(1+2cos2x+cos²2x)
=(1/4)+(1/2)cos2x+(1/8)(1+cos4x)
=(3/8)+(1/2)cos2x+(1/8)cos4x
∫cos⁴xdx
=∫[(3/8)+(1/2)cos2x+(1/8)cos4x]dx
=(3/8)x+(1/4)sin2x+(1/32)sin4x+C
∫1/(sinx+cosx)dx
=∫1/sin(x+π/4)dx
=∫csc(x+π/4)dx
=ln(csc(x+π/4)-cot(x+π/4))+C