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
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
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:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
`∫ (1)/(sin^2 x cos^2 x) dx`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
