Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
` Note: "Here, we are considering "log x as log_e x`
\[\text{Let I} = \int\frac{\sec x}{\log \left( \sec x + \tan x \right)}dx\]
\[\text{Putting} \log \left( \sec x + \tan x \right) = t\]
\[ \Rightarrow \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \frac{dt}{dx}\]
\[ \Rightarrow \sec x\frac{\left( \sec x + \tan x \right)}{\sec x + \tan x} = \frac{dt}{dx}\]
\[ \Rightarrow \text{sec x dx} = dt\]
\[ \therefore I = \int\frac{dt}{t}\]
\[ = \text{log}\left| t \right| + C\]
\[ = \text{log}\left| \text{log }\left( \sec x + \tan x \right) \right| + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
`∫ (1)/(sin^2 x cos^2 x) dx`
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
