tan-3-x-sec-3-x-dx-

Question Number 138009 by bobhans last updated on 09/Apr/21

\int \tan^{3} (x) \sqrt{\sec^{3} (x)} d x = ?

Answered by EDWIN88 last updated on 09/Apr/21

E = \int \tan^{3} (x) \sec (x) \sqrt{\sec (x)} d x

= \int \tan (x) \sec (x) (\sec^{2} (x) - 1) \sqrt{\sec (x)} d x

l e t t i n g ℓ = \sqrt{\sec (x)} \Rightarrow 2 ℓ d ℓ = \sec (x) \tan (x) d x

E = \int (ℓ^{4} - 1) ℓ (2 ℓ d ℓ)

= 2 \int (ℓ^{5} - ℓ^{2}) d ℓ

= 2 (\frac{ℓ^{6}}{6} - \frac{ℓ^{3}}{3}) + c = 2 (\frac{ℓ^{6}}{6} - \frac{2 ℓ^{3}}{6}) + c

= \frac{1}{3} ℓ^{3} (ℓ^{3} - 2) + c = \frac{1}{3} \sqrt{\sec^{3} (x)} (\sqrt{\sec^{3} (x)} - 2) + c

Answered by liki last updated on 09/Apr/21

Commented by liki last updated on 09/Apr/21

help me pls this question

Commented by mr W last updated on 25/Apr/21

y o u s h o u l d p o s t y o u r q u e s t i o n a s n e w

q u e s t i o n, n o t a s a n s w e r o r c o m m e n t

t o o t h e r q u e s t i o n s!

r e a d h e r e :

Commented by mr W last updated on 25/Apr/21

Commented by mr W last updated on 25/Apr/21

p l e a s e a n s w e r m e i f y o u h a v e

u n d e r s t o o d . i f y o u {d o n}^{'} t f o l l o w t h e

r u l e s o f t h e f o r u m, y o u m a y p e r h a p s

g e t n o h e l p .

i^{'} m j u s t t r y i n g t o h e l p y o u .

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