first divide both num and denom by cos square x
then tanx will cum put tanx=u this is in rd sharma the proof is given
- 1
integration of dx/(asinx+bcosx) can be evaluated by putting a= rcos#.........(1) and b=rsin#.......(2)
so that rsquare cos square# + r square sin square# = a square +b square
r square (cos square# + sin square#) = asquare + bsquare
r = squareroot (asquare + bsquare)
and dividing equation 2 by equation 1 we get
tan# = b by a
this implies # = taninverse(b by a)
now integration dx/ (asinbx+bcosx) = integration dx/ (rcos#sinx + rsin#cosx)
= 1 by r integration dx/ (cos#sinx + sin#cosx)
= 1 by r integration dx/(sin(x+#))
= 1 by r integration cosec(x+#)dx
= 1 by r ln [tan((x+#)/2)] +c
=1 by [square root (a square+b square) ln [tan[{x+tan inverse (b bya)}by2]+c
- 2
integration of dx/(asinx+bcosx) can be evaluated by putting a= rcos#.........(1) and b=rsin#.......(2)
so that rsquare cos square# + r square sin square# = a square +b square
r square (cos square# + sin square#) = asquare + bsquare
r = squareroot (asquare + bsquare)
and dividing equation 2 by equation 1 we get
tan# = b by a
this implies # = taninverse(b by a)
now integration dx/ (asinbx+bcosx) = integration dx/ (rcos#sinx + rsin#cosx)
= 1 by r integration dx/ (cos#sinx + sin#cosx)
= 1 by r integration dx/(sin(x+#))
= 1 by r integration cosec(x+#)dx
= 1 by r ln [tan((x+#)/2)] +c
=1 by [square root (a square+b square) ln [tan[{x+tan inverse (b bya)}by2]+c
- 0
integration of dx/(asinx+bcosx) can be evaluated by putting a= rcos#.........(1) and b=rsin#.......(2)
so that rsquare cos square# + r square sin square# = a square +b square
r square (cos square# + sin square#) = asquare + bsquare
r = squareroot (asquare + bsquare)
and dividing equation 2 by equation 1 we get
tan# = b by a
this implies # = taninverse(b by a)
now integration dx/ (asinbx+bcosx) = integration dx/ (rcos#sinx + rsin#cosx)
= 1 by r integration dx/ (cos#sinx + sin#cosx)
= 1 by r integration dx/(sin(x+#))
= 1 by r integration cosec(x+#)dx
= 1 by r ln [tan((x+#)/2)] +c
=1 by [square root (a square+b square) ln [tan[{x+tan inverse (b bya)}by2]+c
- -3
integration of dx/(asinx+bcosx) can be evaluated by putting a= rcos#.........(1) and b=rsin#.......(2)
so that rsquare cos square# + r square sin square# = a square +b square
r square (cos square# + sin square#) = asquare + bsquare
r = squareroot (asquare + bsquare)
and dividing equation 2 by equation 1 we get
tan# = b by a
this implies # = taninverse(b by a)
now integration dx/ (asinbx+bcosx) = integration dx/ (rcos#sinx + rsin#cosx)
= 1 by r integration dx/ (cos#sinx + sin#cosx)
= 1 by r integration dx/(sin(x+#))
= 1 by r integration cosec(x+#)dx
= 1 by r ln [tan((x+#)/2)] +c
=1 by [square root (a square+b square) ln [tan[{x+tan inverse (b bya)}by2]+c
- 1