Advertisements
Advertisements
प्रश्न
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
उत्तर
I = `int_ (x + sin x)/(1 + cos x ) dx`
= `int_ (x + sin x)/(2cos^2 x/2 dx`
= `∫x/(2cos^2 x/2)dx + ∫(2sin x/2 cos x/2)/(2 cos^2 x/2)dx`
= `(1)/(2) int_ xsec^2 x/2 "dx" + int_ tan x/2 dx`
= `(1)/(2) [ x int_ sec^2 x/2 "dx" - ( int_ (d)/(dx) (x) int_ sec^2 x/2 dx) dx] + int_ tan (x)/(2) dx`
= `(1)/(2) [ x .(tan x/2) /(1/2) - int_ (tan x/2 dx)/(1/2) ] + int_ tan (x)/(2) dx + c`
=` x tan x/2 - int_ tan x / 2 dx + int_ tan x /2 dx + c`
= `x tan x/2 + c`
APPEARS IN
संबंधित प्रश्न
` ∫ 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 integral :-
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Write a value of
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate:
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`