sec-d-

Question Number 96613 by s.ayeni14@yahoo.com last updated on 03/Jun/20

\int \sec θ d θ

Answered by M±th+et+s last updated on 03/Jun/20

\int s e c (θ) \frac{s e c (θ) + t a n (θ)}{s e c (θ) + t a n (θ)} d θ

= l n ∣ s e c (θ) + t a n (θ) ∣ + c

i h a v e m o r e t h a n 9 w a y s t o s o l v e t h i s

i n t e g r a l i w i l l p o s t t h e m l a t e r

Commented by s.ayeni14@yahoo.com last updated on 03/Jun/20

ok sir

Answered by M±th+et+s last updated on 03/Jun/20

= \int s e c (x) + t a n (x) - t a n (x) d x

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

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

l n ∣ 1 + s i n (x) ∣ - l n ∣ c o s (x) ∣ + c

Answered by M±th+et+s last updated on 03/Jun/20

= \int \frac{1}{s i n (\frac{π}{2} - x)} d x

= \int \frac{1}{2 s i n (\frac{π}{4} - x) c o s (\frac{π}{4} - x)} . \frac{{s e c}^{2} (\frac{π}{4} - \frac{x}{2})}{{s e c}^{2} (\frac{π}{4} - \frac{x}{2})} d x

= \frac{1}{2} \int \frac{{s e c}^{2} (\frac{π}{4} - \frac{x}{2})}{t a n (\frac{π}{4} - \frac{x}{2})} d x

= - l n ∣ t a n (\frac{π}{4} - \frac{x}{2}) ∣ + c

Answered by M±th+et+s last updated on 03/Jun/20

\int s e c (x) + t a n (x) - t a n (x) d x

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

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

= - 2 l n ∣ c o s (\frac{x}{2}) - \frac{x}{2} ∣ + l n ∣ c o s (x) ∣ + c

Answered by M±th+et+s last updated on 03/Jun/20

= \int s e c (x) \frac{t a n (x)}{t a n (x)} d x

= \int \frac{s e c (x) t a n (x)}{\sqrt{{s e c}^{2} (x) - 1}} d x

= {c o s h}^{- 1} (s e c (x)) + c

Answered by M±th+et+s last updated on 03/Jun/20

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

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

= {s e c h}^{- 1} (c o s (x)) + c

Answered by M±th+et+s last updated on 03/Jun/20

\int \frac{c s c (x)}{c o t (x)} d x

\int \frac{c s c (x) c o t (x)}{c o t (x)} d x

\int \frac{- 1}{{c s c}^{2} (x) - 1} d (c s c (x))

{c o t h}^{- 1} (c s c (x)) + c

Answered by M±th+et+s last updated on 03/Jun/20

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

\int \frac{{s e c}^{2} (\frac{x}{2})}{1 - {t a n}^{2} (\frac{x}{2})} d x

2 \int \frac{- \frac{1}{2} {s e c}^{2} (\frac{x}{2})}{{t a n}^{2} (\frac{x}{2}) - 1} d x

= 2 {c o t h}^{- 1} (t a n (\frac{x}{2})) + c

Answered by M±th+et+s last updated on 03/Jun/20

\int \frac{c s c (x)}{c o t (x)} d x

\int \frac{{c s c}^{2} (x)}{c s c (x) c o t (x)} d x

\int \frac{{c s c}^{2} (x)}{c o t (x) \sqrt{{c o t}^{2} (x) + 1}} d x

{c s c h}^{- 1} (c o t (x)) + c

Answered by M±th+et+s last updated on 03/Jun/20

\int \frac{{s e c}^{2} (\frac{x}{2})}{1 - {t a n}^{2} (\frac{x}{2})} d x

\int \frac{t a n (\frac{x}{2}) {s e c}^{2} (\frac{x}{2})}{{(\sqrt{1 - {t a n}^{2} (\frac{x}{2})})}^{2} t a n \frac{x}{2}} d x

\int \frac{1}{\sqrt{1 - {t a n}^{2} (\frac{x}{2})} \sqrt{1 - {(\sqrt{1 - {t a n}^{2} \frac{x}{2}})}^{2}}} . \frac{2 t a n \frac{x}{2} {s e c}^{2} \frac{x}{2}}{2 \sqrt{1 - {t a n}^{2} \frac{x}{2}}} d x

= 2 {s e c h}^{- 1} \sqrt{1 - {t a n}^{2} (\frac{x}{2})} + c

Answered by M±th+et+s last updated on 03/Jun/20

\int \frac{s e c (x) t a n (x)}{\sqrt{{s e c}^{2} (x) - 1}} d x = \int \frac{s e c (x) t a n (x)}{\sqrt{s e c (x) - 1} \sqrt{s e c (x) + 1}} d x

= \frac{1}{2} \int \frac{s e c (x) t a n (x)}{\sqrt{\frac{s e c (x) - 1}{2}} \sqrt{\frac{s e c (x) + 1}{2}}} d x

= 2 \int \frac{1}{\sqrt{{(\sqrt{\frac{s e c (x) + 1}{2}})}^{2} - 1}} . \frac{\frac{s e c (x) t a n (x)}{2}}{2 \sqrt{\frac{s e c (x) + 1}{2}}}

= 2 {c o s h}^{- 1} (\sqrt{\frac{s e c (x) + 1}{2}}) + c

Answered by M±th+et+s last updated on 03/Jun/20

\int s e c (x) d x = \int \frac{s e c (x) t a n (x)}{\sqrt{{s e c}^{2} (x) - 1}} d x

= \frac{1}{2} \int \frac{s e c (x) t a n (x)}{\sqrt{\frac{s e c (x) - 1}{2}} \sqrt{\frac{s e c (x) + 1}{2}}} d x

= \int \frac{1}{\sqrt{\frac{s e c (x) - 1}{2} + 1}} . \frac{s e c (x) t a n (x)}{2 \sqrt{\frac{s e c (x) - 1}{2}}} d x

= 2 {s i n h}^{-} (\sqrt{\frac{s e c (x) - 1}{2}}) + c

a n d m y b e t h e r e a r e a n o t h e r w a y s

Answered by Rio Michael last updated on 03/Jun/20

hahahaha great work sir!

but one method left .

\int \sec θ d θ = \int \frac{\sec θ (\sec θ + \tan θ)}{\sec θ + \tan θ} d θ = \int \frac{\sec^{2} θ + \sec θ \tan θ}{\sec θ + \tan θ} d θ

let u = \sec θ + \tan θ \Rightarrow d u = (\sec θ \tan θ + \sec^{2} θ) d θ = (\sec^{2} θ + \sec θ \tan θ) d θ

\Rightarrow \int \sec θ d θ = \int \frac{d u}{u} after substitution = \ln ∣ u ∣ + k

\Rightarrow \int \sec θ d θ = \ln ∣ \sec θ + \tan θ ∣ + k