dx-sec-x-csc-x-

Question Number 82185 by jagoll last updated on 19/Feb/20

\int \frac{d x}{\sec x + c s c x} = ?

Commented by john santu last updated on 19/Feb/20

\sec x + c s c x = \frac{1}{\cos x} + \frac{1}{\sin x}

\frac{\sin x + \cos x}{\sin x \cos x} .

\int \frac{\sin x \cos x}{\sin x + \cos x} d x =

l e t \tan (\frac{x}{2}) = t

Commented by mathmax by abdo last updated on 19/Feb/20

l e t I = \int \frac{d x}{\frac{1}{c o s x} + \frac{1}{s i n x}} \Rightarrow I = \int \frac{c o s x . s i n x}{c o s x + s i n x} d x c h a n g e m e n t

t a n (\frac{x}{2}) = t g i v e I = \int \frac{\frac{1 - t^{2}}{1 + t^{2}} \times \frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}} = \int \frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2} (- t^{2} + 2 t + 1)} d t

= \int \frac{2 t (t^{2} - 1)}{{(t^{2} + 1)}^{2} (t^{2} - 2 t - 1)} d t = \int \frac{2 t^{3} - 2 t}{{(t^{2} + 1)}^{2} (t^{2} - 2 t - 1)} d t

t^{2} - 2 t - 1 = 0 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} a n d t_{2} = 1 - \sqrt{2} l e t d e c o m p o s e

F (t) = \frac{2 t^{3} - 2 t}{{(t^{2} + 1)}^{2} (t - t_{1}) (t - t_{2})} \Rightarrow F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1} + \frac{e t + f}{{(t^{2} + 1)}^{2}}

\Rightarrow \int F (t) d t = a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n (t)

+ \int \frac{e t + f}{{(t^{2} + 1)}^{2}} d t r e s t c a l c u l u s o f c o e f f i c i e n t s \dots b e c o n t i n u e d \dots

Answered by TANMAY PANACEA last updated on 20/Feb/20

\int \frac{s i n x + c o s x}{s i n x c o s x} d x

2 \int \frac{s i n x + c o s x}{1 - (1 - 2 s i n x c o s x)} d x

2 \int \frac{d (s i n x - c o s x)}{1 - {(s i n x - c o s x)}^{2}} [f o r m u l a \int \frac{d a}{1 - a^{2}}]

l n (\frac{s i n x - c o s x + 1}{s i n x - c o s x - 1}) + c

Commented by jagoll last updated on 20/Feb/20

\frac{1}{c s c x + \sec x} = \frac{1}{\frac{1}{\sin x} + \frac{1}{\cos x}}

= \frac{\sin x . \cos x}{\sin x + \cos x} \neq \frac{\sin x + \cos x}{\sin x \cos x}

s o r r y s i r . y o u r a n s w e r n o t c o r r e c t

Commented by TANMAY PANACEA last updated on 20/Feb/20