# ∫ (tanx) / (a+b tan2x) dx , plz answer asap, thank you.

32 Points
6 years ago
First write down the whole fuction using sines and cosines, then you will notice that the you can convert this as
the derivarive of denominator is equal to the numerator.
$I=\int \frac{tan(x)}{a+btan^2(x)} dx = \int \frac{sin(x)cos(x)}{a+(b-a)sin^2(x)} dx$
$=\frac{1}{2(b-a)}\int\frac{d}{dx} ln|a+(b-a)sin^2(x)|dx = \frac{1}{2(b-a)} ln|a+(b-a)sin^2(x)| + c$

Ram Gupta
26 Points
3 years ago
First, write down the whole function using sines and cosines

(int = symbol of integration)
= a/b
b

int           sin(x)/cos(x)
a + b.sin^2(x)/cos^2(x)

int             sin(x).cos(x)
a.cos^2(x) + b.sin^2(x)

Let, a.cos^2(x) + b.sin^2(x) = t    ….........................eq ( i )

a.cos(x)[-sin(x)].dx + b.2sin(x).cos(x).dx = dt

cos(x).sin(x).dx = dt.½.(b – a) ….......................eq ( ii )

Therefore substituting the value of eq (i) and eq (ii) in the above question

int     [1/t.   dt.1/2(b – a)]

1/2(b – a)   int   dt/t      ….......................Since 1/2(b – a) is a constant

1/2(b – a) .  [loge|t|  + c]

1/2(b – a).loge |a.cos^2(x) + b.sin^2(x)| + c  …............ is the answer