`int (dx)/(a cos x+ b sin x)=(1)/(r) In[tan((x+alpha)/(2))]` के रूप का है ।
`alpha` का मान ज्ञात कीजिये ।

Options

A. `"tan"^(-1)((a)/(b))`
B. `"tan"^(-1)((b)/(a))`
C. `"tan"^(-1)((a+b)/(a-b))`
D. `"tan"^(-1)((a-b)/(a+b))`

Answer

A

Explanation

`I= int (dx)/(a cos x b sin x)`
`:. alpha= r sin alpha` तथा `b= r cos alpha` रखने पर
`I= (1)/(r) (dx)/(sin alpha* cos x+ cos alpha* sin x)`
`=(1)/(r) int (dx)/(sin (x+alpha))=(1)/(r) int "cosec" (x+alpha)dx`
`=(1)/(r) In [ "cosec" (x+alpha)-cot (x+alpha)]+C`
`=(1)/(r)In [(1)/(sin(x+alpha))-(cos(x+alpha))/(sin(x+alpha))]+C`
`=(1)/(r)In [(1-cos (x+alpha))/(sin (x+alpha))]+C`
`=(1)/(r)In[(2sin^(2)((x+alpha)/(2)))/(2 sin ((x+alpha)/(2))*((x+alpha)/(2)))]+C`
`=(1)/(r)In [ tan((x+alpha)/(2))]+C`
`alpha= tan^(-1)(a//b)`