Advertisements
Advertisements
Question
Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`
Advertisements
Solution
Let `I==int_0^pie^(2x)sin(pi/2+x)dx`
Integrating by parts, we get
` I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/2int_0^pie^(2x)cos(pi/4+x)dx`
Now, integrating the second term by parts, we get
` =>I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/2{[1/2e^(2x)cos(pi/4+x)_0^pi]+1/2int_0^pi e^(2x)sin(pi/4+x)dx}`
=>`I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/4[e^(2x)cos(pi/4+x)_0^pi]-1/4I`
`=>5/4I=1/2[e^(2x)sin(pi+pi/4)-sin(pi/4)]-1/4[e^(2x)cos(pi+pi/4)-cos(pi/4)]`
`=>5/4I=1/2 |__-e^(2x)xx1/sqrt2-1/sqrt2__|-1/4|__-e^(2pi)xx1/sqrt2-1/sqrt2__|`
`=>5/4I==1/(2sqrt2)e^(2pi)-1/(2sqrt2)+1/(4sqrt2)e^(2pi)+1/(4sqrt2)`
`=>I=-1/(5sqrt2)(e^(2pi)+1)`
APPEARS IN
RELATED QUESTIONS
Integrate the function in `x^2e^x`.
Integrate the function in x log 2x.
Integrate the function in x sec2 x.
Integrate the function in `e^x (1 + sin x)/(1+cos x)`.
Evaluate the following : `int e^(2x).cos 3x.dx`
Evaluate the following: `int logx/x.dx`
Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`
Choose the correct options from the given alternatives :
`int (1)/(cosx - cos^2x)*dx` =
Evaluate the following.
`int "e"^"x" "x - 1"/("x + 1")^3` dx
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
Choose the correct alternative from the following.
`int (1 - "x")^(-2) "dx"` =
Evaluate: Find the primitive of `1/(1 + "e"^"x")`
Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx
`int ("d"x)/(x - x^2)` = ______
State whether the following statement is True or False:
If `int((x - 1)"d"x)/((x + 1)(x - 2))` = A log|x + 1| + B log|x – 2|, then A + B = 1
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
Find `int_0^1 x(tan^-1x) "d"x`
Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.
Solve: `int sqrt(4x^2 + 5)dx`
`int(1-x)^-2 dx` = ______
`int1/sqrt(x^2 - a^2) dx` = ______
Evaluate `int(3x-2)/((x+1)^2(x+3)) 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
`int(xe^x)/((1+x)^2) dx` = ______
Evaluate `int(1 + x + (x^2)/(2!))dx`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Prove that `int sqrt(x^2 - a^2)dx = x/2 sqrt(x^2 - a^2) - a^2/2 log(x + sqrt(x^2 - a^2)) + c`
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^2e^(4x)dx`
