Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Advertisements
उत्तर
I = `int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
= `int (sinx + cosx)/sqrt(sin^2x + cos^2x + 2 sin x cosx) "d"x`
= `int (sin x + cosx)/sqrt((sinx + cosx)^2) "d"x`
= `int 1"d"x`
= x + C
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
sin4 x
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
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:
`(cos 2x+ 2sin^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 (2x)/((x^2 + 1)(x^4 + 4))`dx
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int "dx"/(2sin^2x + 5cos^2x)`
Find `int x^2tan^-1x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int sinx/(3 + 4cos^2x) "d"x` = ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
Using \(\cos^2 x = \frac{1+\cos 2x}{2}\), which expression is equivalent to \(\int \cos^2 x\,dx\)?
What should be done first when evaluating an integral containing a complicated trigonometric expression?
Which procedure is recommended when an identity can simplify a complicated trigonometric expression?
