Type a command or search the notes. Tab cycles completions. Up and down recall history when blank. Escape dismisses, then leaves the prompt.
MATH
[1251]

Examples from Trigonometric and hyperbolic substitutions

All examples from Trigonometric and hyperbolic substitutions.

Back to Trigonometric and hyperbolic substitutions

∫1−x2 dx\int\sqrt{1-x^2}\,dx

The integrand is real on [−1,1][-1,1]. On the interior of this interval, let x=sin⁡θx=\sin\theta, so dx=cos⁡θ dθdx=\cos\theta\,d\theta. On the range of sin⁡−1\sin^{-1}, one has 1−sin⁡2θ=∣cos⁡θ∣=cos⁡θ\sqrt{1-\sin^2\theta}=\lvert\cos\theta\rvert=\cos\theta. Then

I=∫1−sin⁡2θ cos⁡θ dθ=∫cos⁡2θ dθ=θ2+sin⁡2θ4+C=θ2+sin⁡θcos⁡θ2+C=12sin⁡−1x+12x1−x2+C.\begin{aligned} I&=\int\sqrt{1-\sin^2\theta}\,\cos\theta\,d\theta\\ &=\int\cos^2\theta\,d\theta\\ &=\frac{\theta}{2}+\frac{\sin 2\theta}{4}+C\\ &=\frac{\theta}{2}+\frac{\sin\theta\cos\theta}{2}+C\\ &=\frac12\sin^{-1}x+\frac12 x\sqrt{1-x^2}+C. \end{aligned}

The integral ∫cos⁡2θ dθ\int\cos^2\theta\,d\theta is the even-power evaluation. Back-substitution uses sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta=2\sin\theta\cos\theta, θ=sin⁡−1x\theta=\sin^{-1}x, sin⁡θ=x\sin\theta=x and cos⁡θ=1−x2\cos\theta=\sqrt{1-x^2}.

[exam]

  1. [1]:Substitutions for a2±x2\sqrt{a^2\pm x^2} and x2−a2\sqrt{x^2-a^2}
  2. [2]:Both powers even

∫04x216−x2 dx\int_0^4 x^2\sqrt{16-x^2}\,dx

Let x=4sin⁡θx=4\sin\theta, so dx=4cos⁡θ dθdx=4\cos\theta\,d\theta. The limits x=0x=0 and x=4x=4 become θ=0\theta=0 and θ=π2\theta=\frac{\pi}{2}. On [0,π2]\bigl[0,\frac{\pi}{2}\bigr] one has 16−16sin⁡2θ=16(1−sin⁡2θ)=4cos⁡θ\sqrt{16-16\sin^2\theta}=\sqrt{16(1-\sin^2\theta)}=4\cos\theta. Then

I=∫0π/216sin⁡2θ 16−16sin⁡2θ 4cos⁡θ dθ=∫0π/216sin⁡2θ⋅4cos⁡θ⋅4cos⁡θ dθ=162∫0π/2sin⁡2θcos⁡2θ dθ=162∫0π/2(sin⁡2θ2)2 dθ=1624∫0π/21−cos⁡4θ2 dθ=32∫0π/2(1−cos⁡4θ) dθ=32[θ−sin⁡4θ4]0π/2=32(π2−sin⁡2π4)=16π.\begin{aligned} I&=\int_0^{\pi/2}16\sin^2\theta\,\sqrt{16-16\sin^2\theta}\,4\cos\theta\,d\theta\\ &=\int_0^{\pi/2}16\sin^2\theta\cdot 4\cos\theta\cdot 4\cos\theta\,d\theta\\ &=16^2\int_0^{\pi/2}\sin^2\theta\cos^2\theta\,d\theta\\ &=16^2\int_0^{\pi/2}\left(\frac{\sin 2\theta}{2}\right)^2\,d\theta\\ &=\frac{16^2}{4}\int_0^{\pi/2}\frac{1-\cos 4\theta}{2}\,d\theta\\ &=32\int_0^{\pi/2}(1-\cos 4\theta)\,d\theta\\ &=32\left[\theta-\frac{\sin 4\theta}{4}\right]_0^{\pi/2}\\ &=32\left(\frac{\pi}{2}-\frac{\sin 2\pi}{4}\right)\\ &=16\pi. \end{aligned}

The calculation uses sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta=\frac{\sin 2\theta}{2} and sin⁡22θ=1−cos⁡4θ2\sin^2 2\theta=\frac{1-\cos 4\theta}{2}.

[exam]

  1. [1]:Substitutions for a2±x2\sqrt{a^2\pm x^2} and x2−a2\sqrt{x^2-a^2}

∫dx(a2+x2)3/2\int\frac{dx}{(a^2+x^2)^{3/2}}

Assume a>0a>0 and let x=atan⁡θx=a\tan\theta with −π2<θ<π2-\frac{\pi}{2}<\theta<\frac{\pi}{2}, so dx=asec⁡2θ dθdx=a\sec^2\theta\,d\theta and sec⁡θ>0\sec\theta>0. Then

I=∫asec⁡2θ(a2+a2tan⁡2θ)3/2 dθ=1a2∫sec⁡2θ(1+tan⁡2θ)3/2 dθ=1a2∫sec⁡2θsec⁡3θ dθ=1a2∫1sec⁡θ dθ=1a2∫cos⁡θ dθ=1a2sin⁡θ+C=xa2x2+a2+C.\begin{aligned} I&=\int\frac{a\sec^2\theta}{(a^2+a^2\tan^2\theta)^{3/2}}\,d\theta\\ &=\frac{1}{a^2}\int\frac{\sec^2\theta}{(1+\tan^2\theta)^{3/2}}\,d\theta\\ &=\frac{1}{a^2}\int\frac{\sec^2\theta}{\sec^3\theta}\,d\theta\\ &=\frac{1}{a^2}\int\frac{1}{\sec\theta}\,d\theta\\ &=\frac{1}{a^2}\int\cos\theta\,d\theta\\ &=\frac{1}{a^2}\sin\theta+C\\ &=\frac{x}{a^2\sqrt{x^2+a^2}}+C. \end{aligned}

The identity sec⁡2θ=tan⁡2θ+1\sec^2\theta=\tan^2\theta+1 gives (1+tan⁡2θ)3/2=sec⁡3θ(1+\tan^2\theta)^{3/2}=\sec^3\theta because sec⁡θ>0\sec\theta>0 on the chosen branch. Also cos⁡θ=a/x2+a2\cos\theta=a/\sqrt{x^2+a^2}, so sin⁡θ=tan⁡θcos⁡θ=x/x2+a2\sin\theta=\tan\theta\cos\theta=x/\sqrt{x^2+a^2} even when x<0x<0.

[exam]

  1. [1]:Substitutions for a2±x2\sqrt{a^2\pm x^2} and x2−a2\sqrt{x^2-a^2}
  2. [2]:A repeated irreducible quadratic

∫dx4+x2\int\frac{dx}{\sqrt{4+x^2}}

Let x=2sinh⁡θx=2\sinh\theta, so dx=2cosh⁡θ dθdx=2\cosh\theta\,d\theta. Then 4+x2=4+4sinh⁡2θ=2cosh⁡θ\sqrt{4+x^2}=\sqrt{4+4\sinh^2\theta}=2\cosh\theta, using cosh⁡2θ−sinh⁡2θ=1\cosh^2\theta-\sinh^2\theta=1 with cosh⁡θ>0\cosh\theta>0. Hence

I=∫2cosh⁡θ2cosh⁡θ dθ=θ+C=sinh⁡−1 ⁣(x2)+C.I=\int\frac{2\cosh\theta}{2\cosh\theta}\,d\theta=\theta+C=\sinh^{-1}\!\left(\frac{x}{2}\right)+C.

The trigonometric route on the same row, x=2tan⁡θx=2\tan\theta, leads to ∫sec⁡θ dθ\int\sec\theta\,d\theta instead.

[routine]

  1. [1]:Substitutions for a2±x2\sqrt{a^2\pm x^2} and x2−a2\sqrt{x^2-a^2}

On this page

∫1−x2 dx\int\sqrt{1-x^2}\,dx∫04x216−x2 dx\int_0^4 x^2\sqrt{16-x^2}\,dx∫dx(a2+x2)3/2\int\frac{dx}{(a^2+x^2)^{3/2}}∫dx4+x2\int\frac{dx}{\sqrt{4+x^2}}