Advertisements
Advertisements
प्रश्न
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Advertisements
उत्तर
Let `I = int ((cos 2x - cos 2alpha)/(cos x - cos alpha)) dx`
`= int ((2 cos^2 x - 1) - (2 cos^2 alpha - 1))/(cos x - cos alpha) dx`
`= int (2(cos^2 x - cos^2 alpha))/ (cos x - cos alpha) dx`
`= int ((2 cos^2 x - 1) - (2 cos^2 alpha - 1))/(cos x - cos alpha) dx`
`= 2 int (cos x + cos alpha) dx`
`= 2 int (cos x) dx + int (2 cos alpha) dx`
`= 2 sin + 2x cos alpha + C`
`= 2 (sin x + x cos alpha) + C`
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
`(sin^3 x + cos^3 x)/(sin^2x cos^2 x)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
Find the integrals of the function:
sin−1 (cos x)
Find `int dx/(x^2 + 4x + 8)`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find `int_ (log "x")^2 d"x"`
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `int sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
`int sinx/(3 + 4cos^2x) "d"x` = ______.
Which identity is used for \(\cos^2 x\)?
Using \(\cos^2 x = \frac{1+\cos 2x}{2}\), which expression is equivalent to \(\int \cos^2 x\,dx\)?
What is \(\int \cos^2 x\,dx\)?
Applying the product-to-sum identity to \(\sin 2x\cos 3x\) gives which expression?
Which procedure is recommended when an identity can simplify a complicated trigonometric expression?
What must always be added to an indefinite integral?
