-dx-cos x - -3sin x equals

I=∫dx/cos x+√(3)sin x ⇒ I=∫dx/2(1/2cos x+√(3)/2sin x) ⇒ I=1/2∫dx/sinπ/6cos x+cosπ/6sin x ⇒ I=1/2∫dx/sin(π/6+x) ⇒ I=1/2∫(π/6+x)dx ⇒ I=1/2log{(π/6+x) (π/6+x) }+c ...

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Byju's Answer

Standard XI

Mathematics

Integration by Substitution

∫dx/cos x + √...

Question

$\int \frac{d x}{cos x + \sqrt{3} sin x}$ equals

log tan (\frac{x}{2} + \frac{π}{12}) + c

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log tan (\frac{x}{2} - \frac{π}{12}) + c

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\frac{1}{2} log tan (\frac{x}{2} + \frac{π}{12}) + c

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\frac{1}{2} log tan (\frac{x}{2} - \frac{π}{12}) + c

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Solution

The correct option is C $\frac{1}{2} log tan (\frac{x}{2} + \frac{π}{12}) + c$
$I = \int \frac{d x}{cos x + \sqrt{3} sin x} \Rightarrow I = \int \frac{d x}{2 (\frac{1}{2} cos x + \frac{\sqrt{3}}{2} sin x)} \Rightarrow I = \frac{1}{2} \int \frac{d x}{sin \frac{π}{6} cos x + cos \frac{π}{6} sin x} \Rightarrow I = \frac{1}{2} \int \frac{d x}{sin (\frac{π}{6} + x)} \Rightarrow I = \frac{1}{2} \int csc (\frac{π}{6} + x) d x \Rightarrow I = \frac{1}{2} log {csc (\frac{π}{6} + x) - cot (\frac{π}{6} + x)} + c \Rightarrow I = \frac{1}{2} log {\frac{1 - cos (\frac{π}{6} + x)}{sin (\frac{π}{6} + x)}} + c \Rightarrow I = \frac{1}{2} log {\frac{2 {sin}^{2} (\frac{π}{12} + \frac{x}{2})}{2 sin (\frac{π}{12} + \frac{x}{2}) cos (\frac{π}{12} + \frac{x}{2})}} + c \Rightarrow I = \frac{1}{2} log {tan (\frac{π}{12} + \frac{x}{2})} + c$

Suggest Corrections

∫dxcosx+√3sinx equals

$\int \frac{d x}{cos x + \sqrt{3} sin x}$ equals