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∫ Sec 2 X 1 − Tan 2 X D X - Mathematics

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प्रश्न

\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
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उत्तर

`  ∫   sec^2 x /{1- tan^2 x }` dx
\[\text{let }\tan x = t\]
\[ \Rightarrow \sec^2 \text{ x dx }= dt\]
Now, `  ∫   sec^2 x /{1- tan^2 x }` dx

\[ = \int\frac{dt}{1 - t^2}\]
\[ = \frac{1}{2} \text{ log } \left| \frac{1 + t}{1 - t} \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| \frac{1 + \tan x}{1 - \tan x} \right| + C\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.16 [पृष्ठ ९०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.16 | Q 1 | पृष्ठ ९०

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