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

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