Question

Integration:-

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

Answer 1

Solution:-

$\bold \green{\displaystyle {\int \cfrac{\sec^2x}{\cosec^2x}dx}}$

$\large {\int {\frac{1}{ \frac{ \cos^2x }{ \frac{1}{ \sin^2x } }dx}}}$

$\large { \int( \frac{1 }{ { \cos }^{2}x } \times \frac{ { \sin }^{2}x }{1})dx }$

$\large{ \int \frac{ \sin^{2}x }{ \cos^{2}x } dx}$

$\large{ \int \tan^{2}x \: dx }$

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${ \fbox{ \sf{\green{we \: know \: that }}}}$

$1 + { \tan }^{2} xθ  = { \sec }^{2}θ$

${ \tan }^{2} θ  = { \sec }^{2} θ  - 1$

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

$= \int { \sec }^{2} x \: dx - \int1dx$

$= \int { \sec }^{2} x \: dx - \int1dx$

$= \int { \sec }^{2}x \: dx - \int {x}^{0} dx$

$= \tan x - \frac{ {x}^{0} + 1 }{0 + 1} + c$

$= \tan x - \frac{ {x}^{1} }{1} + c$

$= \bold {\tan x - x + c}$

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Hope the above concept is clear

Answer 2

Answer:

$\huge\blue{\mid{\fbox{\tt{αηsωєя✿࿐}}\mid}}$

Please refer to the above attachment