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Quiz 2 Solutions

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Question 1

Which integral(s) will give the area of a circle of radius aa?

  1. ∫0a(a2−x2) dx\displaystyle \int_0^a \p{a^2 - x^2} \,\diff{x}
  2. ∫02πa2−x2 dx\displaystyle \int_0^{2\pi} \sqrt{a^2 - x^2} \,\diff{x}
  3. 4∫0aa2−x2 dx\displaystyle 4\int_0^a \sqrt{a^2 - x^2} \,\diff{x}
  4. 4∫a0a2−x2 dx\displaystyle 4\int_a^0 \sqrt{a^2 - x^2} \,\diff{x}
Solution.

(3) is the only solution.

(1) is a polynomial, so this integral will end up being something like (fraction)⋅a2\p{\text{fraction}} \cdot a^2, so it won't be able to get πa2\pi a^2 from this.

(2) doesn't work in general either. For example, if a=1a = 1, then you end up integrating over a region where the function is undefined ([1,2π]\br{1, 2\pi}).

(3) does work, and there are two ways to see that. One way is to recall that x2+y2=a2x^2 + y^2 = a^2 is the circle of radius aa, so if you try solving for yy, you get y=a2−x2y = \sqrt{a^2 - x^2} which is the top-half of the circle. So, if you integrate from 00 to aa, then you get 14\frac{1}{4} of the area of the circle.

The other way is to do the substitution x=asin⁡θx = a\sin\theta directly: dx=acos⁡θ dθ\diff{x} = a\cos\theta \,\diff\theta and a2−x2=acos⁡θ\sqrt{a^2 - x^2} = a\cos\theta, which means

4∫0aa2−x2 dx=4∫0π/2acos⁡θ⋅acos⁡θ dθ=4a2∫0π/2cos⁡2θ dθ=4a2∫0π/21−cos⁡2θ2 dθ=2a2(x−sin⁡2θ2)∣0π/2=2a2π2=πa2.\begin{aligned} 4\int_0^a \sqrt{a^2 - x^2} \,\diff{x} &= 4\int_0^{\pi/2} a\cos\theta \cdot a\cos\theta \,\diff\theta \\ &= 4a^2 \int_0^{\pi/2} \cos^2\theta \,\diff\theta \\ &= 4a^2 \int_0^{\pi/2} \frac{1 - \cos2\theta}{2} \,\diff\theta \\ &= \left. 2a^2 \p{x - \frac{\sin2\theta}{2}} \right\rvert_0^{\pi/2} \\ &= 2a^2 \frac{\pi}{2} \\ &= \pi a^2. \end{aligned}

(4) doesn't work since you'll end up getting −πa2-\pi a^2 instead of πa2\pi a^2.

Question 2

To evaluate the integral ∫xsin⁡xcos⁡x dx\displaystyle \int x \sin{x} \cos{x} \,\diff{x} using integration by parts, the most convenient choice is

  1. u=sin⁡xu = \sin{x}, dv=xcos⁡x dx\diff{v} = x \cos{x} \,\diff{x}
  2. u=sin⁡2xu = \sin{2x}, dv=x dx\diff{v} = x \,\diff{x}
  3. u=xsin⁡xu = x \sin{x}, dv=cos⁡x dx\diff{v} = \cos{x} \,\diff{x}
  4. u=xu = x, dv=sin⁡2x dx\diff{v} = \sin{2x} \,\diff{x}
  5. u=xu = x, dv=cos⁡2x dx\diff{v} = \cos{2x} \,\diff{x}
Solution.

(4) is the solution.

It should (hopefully) be clear that u=xu = x and du=sin⁡xcos⁡x dx\diff{u} = \sin{x} \cos{x} \,\diff{x} works. All you have to do then is remember that 12sin⁡2x=sin⁡xcos⁡x\frac{1}{2}\sin{2x} = \sin{x} \cos{x}, which means that (4) is the right answer.

Question 3

For which of the following integrals will reduction formulas be needed, i.e., where substitution does not work?

  1. ∫sin⁡xcos⁡x dx\displaystyle \int \sin{x} \cos{x} \,\diff{x}
  2. ∫sin⁡7xcos⁡2x dx\displaystyle \int \sin^7{x} \cos^2{x} \,\diff{x}
  3. ∫sin⁡6xcos⁡2x dx\displaystyle \int \sin^6{x} \cos^2{x} \,\diff{x}
  4. ∫sin⁡5x dx\displaystyle \int \sin^5{x} \,\diff{x}
Solution.

(3) is the solution.

Substitution will always work if either the power of sin⁡x\sin{x} or the power of cos⁡x\cos{x} is odd, since you can use the identity sin⁡2x+cos⁡2x=1\sin^2{x} + \cos^2{x} = 1 repeatedly and use u=sin⁡xu = \sin{x} or u=cos⁡xu = \cos{x} to get a polynomial. The only integral where the powers are both even is (3), so you would need a reduction formula for that one.

Question 4

Evaluate the integral ∫cosh⁡−1(3x)9x2−1 dx\displaystyle \int \frac{\cosh^{-1}\p{3x}}{\sqrt{9x^2 - 1}} \,\diff{x}.

Solution.

If u=cosh⁡−1(3x)u = \cosh^{-1}\p{3x}, then

du=39x2−1 dx,\diff{u} = \frac{3}{\sqrt{9x^2 - 1}} \,\diff{x},

so

∫cosh⁡−1(3x)9x2−1=13∫u du=16u2+C=16(cosh⁡−1(3x))2+C.\int \frac{\cosh^{-1}\p{3x}}{\sqrt{9x^2 - 1}} = \frac{1}{3} \int u \,\diff{u} = \frac{1}{6} u^2 + C = \boxed{\frac{1}{6} \p{\cosh^{-1}\p{3x}}^2 + C}.

Question 5

Find I=∫sin⁡xcos⁡2x+5cos⁡x+6 dx\displaystyle I = \int \frac{\sin{x}}{\cos^2{x} + 5\cos{x} + 6} \,\diff{x}.

Solution.

If you use the substituion u=cos⁡xu = \cos{x}, then du=−sin⁡x dx\diff{u} = -\sin{x} \,\diff{x}, which gives you

I=−∫duu2+5u+6.I = -\int \frac{\diff{u}}{u^2 + 5u + 6}.

We can do this by partial fractions, which gives you

1u2+5u+6=1(u+2)(u+3)=1u+2−1u+3.\frac{1}{u^2 + 5u + 6} = \frac{1}{\p{u+2}\p{u+3}} = \frac{1}{u + 2} - \frac{1}{u + 3}.

So,

I=−∫1u+2−1u+3 du=−ln⁡∣u+2∣+ln⁡∣u+3∣+C=−ln⁡(cos⁡x+2)+ln⁡(cos⁡x+3)+C.\begin{aligned} I &= -\int \frac{1}{u + 2} - \frac{1}{u + 3} \,\diff{u} \\ &= -\ln\abs{u + 2} + \ln\abs{u + 3} + C \\ &= \boxed{-\ln\p{\cos{x} + 2} + \ln\p{\cos{x} + 3} + C}. \end{aligned}

We can get rid of the absolute values since cos⁡x+2≥0\cos{x} + 2 \geq 0 and cos⁡x+3≥0\cos{x} + 3 \geq 0.