Calculus AP

hi again
im really need help
TextBook: James Stewart:Essential Calculus, page 311. Here the problem #27: First make a substitution and then use integration by parts to evaluate the integral.
Integral from sqrt(pi/2) TO sqrt(pi)of θ^3 cos(θ^2)dθ.

i did my problem: let t=θ^2 then dt= 2θdθ.

∫θ^3cos(θ^2)dθ = ∫1/2 t cos t dt

=1/2∫tcostdt -> let u=t, dv=cost dtthen du = 1, v= sin t

=1/2[tsint-∫sint dt]
=1/2 tsint + 1/2 cos t + c.. reminder t=θ^2
=1/2 θ^2 sin(θ^2)+ 1/2 cos(θ^2) + c

so i got stuck and don't know how to solve with sqrt(pi) and sqrt(pi/2)

asked by Vicky
  1. Up to 1/2∫tcostdt, you're OK.

    After that, you need to integrate by parts.

    I=(1/2)∫t cos(t)dt
    =(1/2)[t sin(t) - ∫sin(t)]
    =(1/2)[t sin(t) + cos(t)

    To evaluate the definite integral, remember to adjust the limits accordingly, i.e.
    from x^2 to t

    I get -(π+2)/4

    posted by MathMate

Respond to this Question

First Name

Your Response

Similar Questions

  1. Math - Calculus

    The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below. |
  2. Calculus II

    Evaluate using u-substitution: Integral of: 4x(tan(x^2))dx Integral of: (1/(sqrt(x)*x^(sqrt(x))))dx Integral of: (cos(lnx)/x)dx
  3. Pre-Calculus

    If θ represents an angle such that sin2θ = tanθ - cos2θ, then sin θ - cosθ = A. -√2 B. 0 C. 1 D. 2√2 What equation can I use to solve this problem?
  4. Math

    Evaluate the integral from [sqrt(pi/2), sqrt(pi)] of x^3*cos(x^2) by first making a substitution and then using integration by parts. I let u = x^2 and du= 2x dx but then it doesn't equal that in the equation?
  5. Calculus URGENT test tonight

    Integral of: __1__ (sqrt(x)+1)^2 dx The answer is: 2ln abs(1+sqrt(x)) + 2(1+sqrt(X))^-1 +c I have no clue why that is! Please help. I used substitution and made u= sqrt(x)+1 but i don't know what happened along the way! Your first
  6. Math - Trig Substitution

    How can I solve the integral of x^3√(9-x^2) dx using trigonometric substitution? ? ∫ x^3√(9-x^2) dx So then I know that x = 3sinθ dx = 3cosθdθ When I substitute, it becomes ∫ (3sinθ)^3 *
  7. Calculus

    Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares, (b) semicircles, and (c) equilateral triangles. for y=x^2,
  8. calculus-integration!

    should i use substitution?? if yes how should should i use it? plz i need some directions? k plz someone? far i used trig. substitution. i got a=8, so i used x=asin(è)so according to this substitution i got x=8sin(è) and
  9. Trigonometry

    There are four complex fourth roots to the number 4−4√3i. These can be expressed in polar form as z1=r1(cosθ1+isinθ1) z2=r2(cosθ2+isinθ2) z3=r3(cosθ3+isinθ3) z4=r4(cosθ4+isinθ4),
  10. Calculus

    I have two questions, because I'm preparing for a math test on monday. 1. Use the fundamental theorem of calculus to find the derivative: (d/dt) the integral over [0, cos t] of (3/5-(u^2))du I have a feeling I will be able to find

More Similar Questions