Question

$\ \textless \ br /\ \textgreater \ {\displaystyle {\int \cfrac{\sec^2x}{\cosec^2x}dx}}$


Please solve this



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Answer 1

Answer:

tanx - x + c

Explanation:

Formulae used:

sec²x - tan²x = 1

∫sec²x dx = tanx + c

Solution:

sec²x / cosec² x = (1/cos² x)/ (1/ sin² x) = sin²x/cos²x

= tan² x = sec²x - 1

Therefore,

∫(sec²x - 1)dx = ∫sec²x dx - ∫dx = tanx - x + c

where, c is a constant

Answer 2

${\displaystyle{\int \dfrac{\sec^2x}{\cosec^2x}dx}}$

${\displaystyle{\int \dfrac{\sin^2x}{\cos^2x}dx}}$

${\displaystyle{\int {\tan^2x•dx}}}$

${\displaystyle{\boxed{\left(1+\tan^2x=\sec^2x\right)}}}$

${\displaystyle{\int (\sec^2x-1)dx}}$

${\displaystyle{\int \sec^2x \int 1•dx}}$

${\displaystyle{\tan x-x+c}}$