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Calculus/Calculus-2 help- trigonometric integrals under principles of integral evaluation chapter

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Question
Hello, Good evening, sir. I have trouble solving this three problems.
1. Rewrite sin x + cos x in the form of A sin(x+0/) And use your results together with previous exercise to evaluate
Integration of dx/sin x + cos x    .....
The previous exercise is integration of csc x dx = -ln|csc x + cot x| + c & prove integration of csc x dx= ln|tan(x/2)|+c   ((((((0/= angle= fay,, becoz I couldn' find the symbol on keyboard)))))

Please help me sir thanks

Answer
Ok Nayan, 1st of all let's write sin(x)+cos(x) as Asin(x+Ø).
You must agree that sin(x)+cos(x)=√2[(1/√2)sin(x)+(1/√2)cos(x)].
It's the same, it's only a mathmatical trick.
Now, we also know that 1/√2 is exactly sin(45°)=cos(45°). That means:
√2[(1/√2)sin(x)+(1/√2)cos(x)]=√2[sin(x)cos(45°)+cos(x)sin(45°].
& according the the tregonometric idintity :
sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
We can write : √2[sin(x)cos(45°)+cos(x)sin(45°]=√2sin(x+45°).
A=√2 & Ø=45°.

As for the 2nd part :
∫dx/(sinx+cosx)=∫dx/√2sin(x+45°)=(1/√2)∫dx/sin(x+45°)=
(1/√2)∫sec(x+45°)dx=(1/√2)Ln|sec(x+45°)+tan(x+45°)|.

Alon.

Calculus

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