Question

find the integral : sec^2x/cosec^2x

Answer 1

∫(sec2x​/cosec2x)dx
​
∫(1/cos​​2x)/(1/sin2x)dx

∫(1/cos​2sin) . (sin2x/1)dx

∫tan2x dx

∫(sec2x-1) dx

tanx - x + c

Answer 2

Answer:

$\int \frac{sec^{2}x}{cosec^{2}x}\:dx</p><p>= tanx - x + c$

Step-by-step explanation:

i) $\frac{sec^{2}x}{cosec^{2}x}$

= $\frac{\frac{1}{cos^{2}x}}{\frac{1}{sin^{2}x}}$

=$\frac{sin^{2}x}{cos^{2}x}$

= $tan^{2}x$ ---(1)

Now ,

$\int \frac{sec^{2}x}{cosec^{2}x}\:dx$

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

\* From (1) *\

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

/* Apply the Trigonometric identity:

tan²x = sec²x - 1 */

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

\* Apply sum rule *\

= $tanx - x + c$

\* Since ,

i ) $\int sec^{2}x\: dx = tanx$

ii )$\int 1\:dx = x$ *\

Therefore,

$\int \frac{sec^{2}x}{cosec^{2}x}\:dx</p><p>= tanx - x + c$

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