Turn the radical into a trigonometric identity

(a) Use x=tan⁡tx=\tan t, where 0<t<π/20<t<\pi/2, to find ∫1x2x2+1dx\int\frac1{x^2\sqrt{x^2+1}}dx. marks

(b)(i) Write down ddx(cos⁡x3)\frac d{dx}(\cos x^3). mark

(ii) Hence find ∫x5sin⁡x3dx\int x^5\sin x^3dx. marks

[2023/SAJC/Prelim/II/3]

Total marks: 9

Solution

(a)

Trigonometric substitution

Use x=tan⁡tx=\tan t, where 0<t<π20<t<\frac\pi2, to find ∫1x2x2+1 dx\int\dfrac1{x^2\sqrt{x^2+1}}\,dx.

Explore this idea

Hint

  1. dx=sec⁡2t dtdx=\sec^2t\,dt and x2+1=sec⁡t\sqrt{x^2+1}=\sec t.

  2. The integrand reduces to cos⁡tsin⁡2t\frac{\cos t}{\sin^2t}.

Answer · (a)

−1+x2/x+C-\sqrt{1+x^2}/x+C.

The stated tan substitution uses x>0.

Explore the idea

For x > 0

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 2.50
Reset the controls

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.337
Below-axis contribution
0
Signed integral
0.337

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 1.00 to x
0.337
antiderivative − F(1)
0.337
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

The integrand decays quickly; the area to infinity is finite.

Ideas to try
As x→∞x\to\infty the area levels off at 2−1≈0.414\sqrt2-1\approx0.414 more.
(b)(i)

Chain rule

Write down ddx(cos⁡x3)\dfrac d{dx}\left(\cos x^3\right).

Hint

  1. Chain rule with inner function x3x^3.

Answer · (b)(i)

−3x2sin⁡(x3)-3x^2\sin(x^3).

(b)(ii)

Integration by parts using the given derivative

Hence find ∫x5sin⁡x3 dx\int x^5\sin x^3\,dx.

Explore this idea

Hint

  1. Write x5sin⁡x3=x3⋅x2sin⁡x3x^5\sin x^3=x^3\cdot x^2\sin x^3.

  2. Take u=x3u=x^3 and dvdx=x2sin⁡x3\frac{dv}{dx}=x^2\sin x^3, so v=−13cos⁡x3v=-\frac13\cos x^3.

Answer · (b)(ii)

−x3cos⁡(x3)/3+sin⁡(x3)/3+C-x^3\cos(x^3)/3+\sin(x^3)/3+C.

Explore the idea

An even integrand and an odd area

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 0.90
Reset the controls

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.041
Below-axis contribution
0
Signed integral
0.041

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.00 to x
0.041
antiderivative − F(0)
0.041
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

For this even integrand, the signed running integral from 0 is odd: G(−x)=−G(x)G(-x)=-G(x). The signed integrals over symmetric intervals [−x,0][-x,0] and [0,x][0,x] are equal.

Ideas to try
The integrand is even: is the area function odd?