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)`
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
sin x sin 2x sin 3x
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
sin4 x
Find the integrals of the function:
cos4 2x
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:
`1/(sin xcos^3 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
`int (sin^2x - cos^2 x)/(sin^2 x cos^2 x) dx` is equal to ______.
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
Find `int dx/(x^2 + 4x + 8)`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) 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 sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"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`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
