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Question

Integrate dx/sinx+√3cosx

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Solution

I=∫ dx/(√3cosx + sinx )

just multiply and divide by 2

so , 2I=∫dx/ ((√3÷2)cosx+ (1÷2)sinx)

2I=∫dx/(cosx * cos 30 + sin30 *sinx)

because sin 30=1/2 and
cos 30= √3/2

2I=∫dx/cos(x-30)

because cosa*cosb+sina*sinb= cos(a-b)

2I=∫sec(x-30) dx

because 1/cosa = seca

I=1/2∫sec(x-30) dx

I= ½ Log[ | tan(x-30) + sex (x-30 ) | ] + C

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