请问这一题怎么解，帮忙解答一下过程。
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∫[1/(sinx+cosx)]dx
=∫[1/√2sin(x+π/4)]d(x+π/4)
=(√2/2)∫[1/sin(x+π/4)]d(x+π/4)
令x+π/4=t，则原式=(√2/2)∫(1/sint)dt
对于∫(1/sint)dt=∫[(sin²(t/2)+cos²(t/2))/(2sin(t/2)cos(t/2))]dt
=∫[(sin²(t/2)+cos²(t/2))/(sin(t/2)cos(t/2))]d(t/2)
=∫[sin(t/2)/cos(t/2)]d(t/2)+∫[cos(t/2)/sin(t/2)]d(t/2)
=-∫[1/cos(t/2)]d[cos(t/2)]+∫[1/(sin(t/2))]d[sin(t/2)]
=-ln|cos(t/2)|+ln|sin(t/2)|+C
=ln|tan(t/2)|+C
代入得到：原式=(√2/2)ln|tan[(x+π/4)/2]|+C
=∫[1/√2sin(x+π/4)]d(x+π/4)
=(√2/2)∫[1/sin(x+π/4)]d(x+π/4)
令x+π/4=t，则原式=(√2/2)∫(1/sint)dt
对于∫(1/sint)dt=∫[(sin²(t/2)+cos²(t/2))/(2sin(t/2)cos(t/2))]dt
=∫[(sin²(t/2)+cos²(t/2))/(sin(t/2)cos(t/2))]d(t/2)
=∫[sin(t/2)/cos(t/2)]d(t/2)+∫[cos(t/2)/sin(t/2)]d(t/2)
=-∫[1/cos(t/2)]d[cos(t/2)]+∫[1/(sin(t/2))]d[sin(t/2)]
=-ln|cos(t/2)|+ln|sin(t/2)|+C
=ln|tan(t/2)|+C
代入得到：原式=(√2/2)ln|tan[(x+π/4)/2]|+C
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