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प्रश्न
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
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उत्तर
Let I = \[\int\frac{dx}{\sin^2 x \cos^2 x}\]
Dividing the numerator and denominator by cos4 x, we get:
I = \[\int\frac{se c^2 x \cdot se c^2 x}{\tan^2 x}dx\]
\[\int\frac{\left( 1 + \tan^2 x \right) \cdot se c^2 x}{\tan^2 x}dx\]
Put tan x = t
⇒ \[se c^2 xdx = dt\]
∴ I = \[\int\frac{1 + t^2}{t^2}dt\] = \[\int1dt + \int\frac{1}{t^2}dt\]
⇒ I = t −\[\frac{1}{t}\] + C
⇒ I = tan x − cot x + C
∴ \[\int\frac{dx}{\sin^2 x \cos^2 x}\] = tan x − cot x + C
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