Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\sec x \cdot \log \left( \text{sec x} + \text{tan x} \right) dx\]
\[ \text{Let log} \left( \sec x + \tan x \right) = t\]
\[ \Rightarrow \frac{\left( \sec x \tan x + \sec^2 x \right)}{\left( \sec x + \tan x \right)} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{\sec x \left( \sec x + \tan x \right)}{\left( \sec x + \tan x \right)} dx = dt\]
\[Now, \int\sec x \cdot \text{log }\left( \sec x + \tan x \right) dx\]
\[ = \ ∫ t . dt\]
\[ = \frac{t^2}{2} + C\]
\[ = \frac{\left[ \text{log} \left( \text{sec x} + \tan x \right) \right]^2}{2} + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate `int_1^3(e^(2-3x)+x^2+1)dx` as a limit of sum.
Evaluate the definite integral:
`int_0^(pi/4) (sin x + cos x)/(9+16sin 2x) dx`
Prove the following:
`int_0^1 xe^x dx = 1`
Prove the following:
`int_(-1)^1 x^17 cos^4 xdx = 0`
Prove the following:
`int_0^(pi/4) 2 tan^3 xdx = 1 - log 2`
Prove the following:
`int_0^1sin^(-1) xdx = pi/2 - 1`
`int dx/(e^x + e^(-x))` is equal to ______.
Choose the correct answers The value of `int_0^1 tan^(-1) (2x -1)/(1+x - x^2)` dx is
(A) 1
(B) 0
(C) –1
(D) `pi/4`
Evaluate the following integral:
Evaluate the following integrals as limit of sums:
\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]
Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.
Solve: (x2 – yx2) dy + (y2 + xy2) dx = 0
Evaluate the following as limit of sum:
`int _0^2 (x^2 + 3) "d"x`
Evaluate the following as limit of sum:
`int_0^2 "e"^x "d"x`
Evaluate the following:
`int_0^2 ("d"x)/("e"^x + "e"^-x)`
Evaluate the following:
`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))` (Hint: Let x = sin θ)
The value of `int_(-pi)^pi sin^3x cos^2x "d"x` is ______.
The limit of the function defined by `f(x) = {{:(|x|/x",", if x ≠ 0),(0",", "otherwisw"):}`
What is the derivative of `f(x) = |x|` at `x` = 0?
`lim_(x -> 0) (xroot(3)(z^2 - (z - x)^2))/(root(3)(8xz - 4x^2) + root(3)(8xz))^4` is equal to
