Advertisements
Advertisements
Question
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Advertisements
Solution
`I = int sec^2 x /sqrt(tan^2 x+4) dx.`
`"Let" tan x = t`
`sec^2 xdx = dt`
So, `I = int dt/sqrt(t^2 + 4)`
or, `I = int dt/sqrt(t^2 + 2^2)`
Since, we Know
`int dx/sqrt(x^2 + a^2) = "In" |x + sqrt(a^2 + x^2)| + "C"`
`I = "In" |t + sqrt(t^2 + 4)| + "C"`
`i.e`
`I = "In"|tanx + sqrt(tan^2 x + 4)| + "C"`
APPEARS IN
RELATED QUESTIONS
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
`cos x/(1 + cos x)`
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−1 (cos x)
Find the integrals of the function:
`1/(cos(x - a) cos(x - b))`
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
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 .
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
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.`
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
What does integration using trigonometric identities mean?
Which identity is used for \(\sin^3 x\)?
After a trigonometric expression has been simplified, how should it be integrated?
