Question

红线这一步骤是怎么化简得出来的呢？谢谢解答。

红线这一步骤是怎么化简得出来的呢？谢谢解答。谢谢解答

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Answer 1

1/(sint+cost)
=1/[√2.cos(t-π/4)]
=(1/√2).sec(t+π/4)
∫(0->π/4) dt/(sint+cost)
=(1/2)[ln|sec(t-π/4) +tan(t-π/4)|]|(0->π/4)