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sinxsinxcosxdx

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a
12lnsinx−cosx+x2+c
b
12lnsinx+cosx+c
c
12lnsinx+2cosx+c
d
None of these

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detailed solution

Correct option is A

Let Nr=Addx(Dr)+B(Dr)sin⁡x=Addx(sin⁡x−cos⁡x)+B(sin⁡x−cos⁡x)sin⁡x=A(cos⁡x+sin⁡x)+B(sin⁡x−cos⁡x)0.cos⁡x+1⋅sin⁡x=(A−B)cos⁡x+(A+B)sin⁡x On comparing both sides A−B=0 and A+B=1 Solving we get A=12 B=12∫sin⁡xsin⁡x−cos⁡xdx=∫12(cos⁡x+sin⁡x)+12(sin⁡x−cos⁡x)sin⁡x−cos⁡xdx=12∫cos⁡x+sin⁡xsin⁡x−cos⁡xdx+12∫sin⁡x−cos⁡xsin⁡x−cos⁡xdx=12loge⁡|sin⁡x−cos⁡x|+12∫1dx=12lnsinx−cosx+12x+c


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