Advertisements
Advertisements
Question
Integrate the following w.r.t.x : sec4x cosec2x
Advertisements
Solution
Let I = `int sec^4x "cosec"^2x*dx`
= `int sec^4x "cosec"^2x* sec^2x*dx`
Put tan x = t
∴ sec2x·dx = d
Also, sec2x cosec2x = (1 + tan2x)(1 + cot2x)
= `(1 + t^2)(1 + 1/t^2)`
= `(1 + t^2)((t^2 + 1)/t^2)`
= `(t^4 + 2t^2 + 1)/t^2`
= `t^2 + 2 + (1)/t^2`
∴ I = `int (t^2 + 2 + 1/t^2)*dt`
= `int t^2*dt + 2 int *dt + int 1/t^2*dt`
= `t^3/(3) + 2t + (t^-1)/((-1)) + c`
= `(1)/(3)tan^3x + 2tanx - (1)/tanx + c`
= `(1)/(3cot^3x) + (2)/(cotx) - cot x + c`.
APPEARS IN
RELATED QUESTIONS
If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:
(A) 0
(B) π
(C) π/2
(D) π/4
Integrate the function in x log x.
Integrate the function in x sec2 x.
Integrate the function in `(xe^x)/(1+x)^2`.
Integrate the function in `e^x (1/x - 1/x^2)`.
Evaluate the following:
`int sec^3x.dx`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Evaluate the following: `int logx/x.dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x: `sqrt(x^2 + 2x + 5)`.
Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
Choose the correct options from the given alternatives :
`int (sin^m x)/(cos^(m+2)x)*dx` =
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`
Evaluate the following.
∫ x log x dx
Evaluate: `int "dx"/("9x"^2 - 25)`
Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`
Evaluate: `int "dx"/(5 - 16"x"^2)`
Evaluate:
∫ (log x)2 dx
`int (cos2x)/(sin^2x cos^2x) "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
`int 1/x "d"x` = ______ + c
`int 1/(x^2 - "a"^2) "d"x` = ______ + c
`int"e"^(4x - 3) "d"x` = ______ + c
`int (x^2 + x - 6)/((x - 2)(x - 1)) "d"x` = x + ______ + c
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
`int cot "x".log [log (sin "x")] "dx"` = ____________.
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
Solve: `int sqrt(4x^2 + 5)dx`
Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.
`intsqrt(1+x) dx` = ______
Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.
Solution: (x2 + y2) dx - 2xy dy = 0
∴ `dy/dx=(x^2+y^2)/(2xy)` ...(1)
Puty = vx
∴ `dy/dx=square`
∴ equation (1) becomes
`x(dv)/dx = square`
∴ `square dv = dx/x`
On integrating, we get
`int(2v)/(1-v^2) dv =intdx/x`
∴ `-log|1-v^2|=log|x|+c_1`
∴ `log|x| + log|1-v^2|=logc ...["where" - c_1 = log c]`
∴ x(1 - v2) = c
By putting the value of v, the general solution of the D.E. is `square`= cx
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
Evaluate the following.
`intx^3e^(x^2) dx`
Evaluate `int(1 + x + x^2/(2!))dx`.
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
Which function is an example under priority \(I\) in the LIATE rule?
Evaluate \[\int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.\]
