Advertisements
Advertisements
प्रश्न
Find the integrals of the function:
tan3 2x sec 2x
Advertisements
उत्तर
Let `I = int tan^3 2x sec 2x dx`
`= int tan^2 2x * tan 2x * sec 2x dx`
`= int (sec^2 2x - 1)* sec 2x tan 2x dx`
Put sec 2x = t
⇒ 2 sec 2x tan 2x dx = dt
∴ `I = 1/2 int (t^2 - 1) dt 1/2 (t^3/3 - 1) + C`
`= 1/6 sec^3 2x - 1/2 sec 2x + 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 3x cos 4x
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
tan4x
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:
`(cos 2x)/(cos x + sin x)^2`
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.
Evaluate `int_0^(3/2) |x sin pix|dx`
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find `int_ (log "x")^2 d"x"`
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
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)`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int tan^2x sec^4 x"d"x`
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
