`int 1/(sin 2x+sin x) dx` का मान ज्ञात कीजिए।

Answer

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Explanation

`int (1)/(sin 2x+sin x) dx=int (1)/(2 sin x cos x+sin x) dx`
`=int (1)/(sin x(2 cos x+1)) dx=int (sin x)/(sin^(2) x(2 cos x+1)) dx`
मानलो `cos x=y`, तो `-sin x dx=dy`
`:. I=int (-1)/((1-y^(2))(2y+1)) dx=int (-1)/((1-y)(1+y)(2y+1)) dy`
`1/((1+y)(1-y)(2y+1)) =(1)/(2(1-y)(2+1))+1/((1+y)(1+1)(-2+1))+(1)/((1-1/2)(1+1/2)(2y+1))`
`=(1)/(6(1-y))-1/(2(1+y))+4/(3(2y+1))`
`:. int (-1)/((1-y)(1+y)(2y+1)) dy = int [1/(2(1+y))-1/(6(1-y))-4/(3(2y+1))] dy`
`=1/2 log |1+y|-1/6 (-1) log |1-y|-4/3 1/2 log |2y+1|+c`
`=1/2 log |1+cos x|+1/6 log |1-cos x|-2/3 log |2 cos x+1|+c`