Trigonometric and hyperbolic substitutions Substitutions that clear a 2 ± x 2 \sqrt{a^2\pm x^2} a 2 ± x 2 x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2
This page clears the quadratic surds a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2
The sign pattern suggests the substitution, but the chosen branch decides which sign of the square root is valid. Right triangles help recover sin θ \sin\theta sin θ cos θ \cos\theta cos θ x x x
Any powers of sine and cosine left over are trigonometric integrals . Other surds may yield to a rationalising substitution that produces a rational function , although not every surd has such a simple substitution.
[ METHOD ] For a > 0 a>0 a > 0 a 2 − a 2 sin 2 θ = a ∣ cos θ ∣ \sqrt{a^2-a^2\sin^2\theta}=a|\cos\theta| a 2 − a 2 sin 2 θ = a ∣ cos θ ∣
Integrand factor Substitutions Identities a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 x = a sin θ x=a\sin\theta x = a sin θ d x = a cos θ d θ dx=a\cos\theta\,d\theta d x = a cos θ d θ x = a tanh θ x=a\tanh\theta x = a tanh θ d x = a sech 2 θ d θ dx=a\operatorname{sech}^2\theta\,d\theta d x = a sech 2 θ d θ cos 2 θ + sin 2 θ = 1 \cos^2\theta+\sin^2\theta=1 cos 2 θ + sin 2 θ = 1 sech 2 θ + tanh 2 θ = 1 \operatorname{sech}^2\theta+\tanh^2\theta=1 sech 2 θ + tanh 2 θ = 1 a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x = a tan θ x=a\tan\theta x = a tan θ d x = a sec 2 θ d θ dx=a\sec^2\theta\,d\theta d x = a sec 2 θ d θ x = a sinh θ x=a\sinh\theta x = a sinh θ d x = a cosh θ d θ dx=a\cosh\theta\,d\theta d x = a cosh θ d θ sec 2 θ = tan 2 θ + 1 \sec^2\theta=\tan^2\theta+1 sec 2 θ = tan 2 θ + 1 cosh 2 θ = 1 + sinh 2 θ \cosh^2\theta=1+\sinh^2\theta cosh 2 θ = 1 + sinh 2 θ x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x = a sec θ x=a\sec\theta x = a sec θ d x = a sec θ tan θ d θ dx=a\sec\theta\tan\theta\,d\theta d x = a sec θ tan θ d θ x ≥ a x\ge a x ≥ a x = a cosh θ x=a\cosh\theta x = a cosh θ d x = a sinh θ d θ dx=a\sinh\theta\,d\theta d x = a sinh θ d θ sec 2 θ = 1 + tan 2 θ \sec^2\theta=1+\tan^2\theta sec 2 θ = 1 + tan 2 θ cosh 2 θ = 1 + sinh 2 θ \cosh^2\theta=1+\sinh^2\theta cosh 2 θ = 1 + sinh 2 θ
For a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 a a a θ \theta θ x x x a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x x x a a a a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x ≥ a x\ge a x ≥ a x x x a a a x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2
The identity sech 2 θ + tanh 2 θ = 1 \operatorname{sech}^2\theta+\tanh^2\theta=1 sech 2 θ + tanh 2 θ = 1 1 − tanh 2 θ = sech 2 θ 1-\tanh^2\theta=\operatorname{sech}^2\theta 1 − tanh 2 θ = sech 2 θ
Writing 1 − sin 2 θ = cos θ \sqrt{1-\sin^2\theta}=\cos\theta 1 − sin 2 θ = cos θ sin − 1 \sin^{-1} sin − 1 cos θ ≥ 0 \cos\theta\ge 0 cos θ ≥ 0 θ \theta θ x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x ≤ − a x\le-a x ≤ − a x = − a cosh θ x=-a\cosh\theta x = − a cosh θ θ ≥ 0 \theta\ge0 θ ≥ 0
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] The integrand is real on [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] x = sin θ x=\sin\theta x = sin θ d x = cos θ d θ dx=\cos\theta\,d\theta d x = cos θ d θ sin − 1 \sin^{-1} sin − 1 1 − sin 2 θ = ∣ cos θ ∣ = cos θ \sqrt{1-\sin^2\theta}=\lvert\cos\theta\rvert=\cos\theta 1 − sin 2 θ = ∣ cos θ ∣ = cos θ
I = ∫ 1 − sin 2 θ cos θ d θ = ∫ cos 2 θ d θ = θ 2 + sin 2 θ 4 + C = θ 2 + sin θ cos θ 2 + C = 1 2 sin − 1 x + 1 2 x 1 − x 2 + 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} I = ∫ 1 − sin 2 θ cos θ d θ = ∫ cos 2 θ d θ = 2 θ + 4 sin 2 θ + C = 2 θ + 2 sin θ cos θ + C = 2 1 sin − 1 x + 2 1 x 1 − x 2 + C . The integral ∫ cos 2 θ d θ \int\cos^2\theta\,d\theta ∫ cos 2 θ d θ even-power evaluation . Back-substitution uses sin 2 θ = 2 sin θ cos θ \sin 2\theta=2\sin\theta\cos\theta sin 2 θ = 2 sin θ cos θ θ = sin − 1 x \theta=\sin^{-1}x θ = sin − 1 x sin θ = x \sin\theta=x sin θ = x cos θ = 1 − x 2 \cos\theta=\sqrt{1-x^2} cos θ = 1 − x 2
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let x = 4 sin θ x=4\sin\theta x = 4 sin θ d x = 4 cos θ d θ dx=4\cos\theta\,d\theta d x = 4 cos θ d θ x = 0 x=0 x = 0 x = 4 x=4 x = 4 θ = 0 \theta=0 θ = 0 θ = π 2 \theta=\frac{\pi}{2} θ = 2 π [ 0 , π 2 ] \bigl[0,\frac{\pi}{2}\bigr] [ 0 , 2 π ] 16 − 16 sin 2 θ = 16 ( 1 − sin 2 θ ) = 4 cos θ \sqrt{16-16\sin^2\theta}=\sqrt{16(1-\sin^2\theta)}=4\cos\theta 16 − 16 sin 2 θ = 16 ( 1 − sin 2 θ ) = 4 cos θ
I = ∫ 0 π / 2 16 sin 2 θ 16 − 16 sin 2 θ 4 cos θ d θ = ∫ 0 π / 2 16 sin 2 θ ⋅ 4 cos θ ⋅ 4 cos θ d θ = 16 2 ∫ 0 π / 2 sin 2 θ cos 2 θ d θ = 16 2 ∫ 0 π / 2 ( sin 2 θ 2 ) 2 d θ = 16 2 4 ∫ 0 π / 2 1 − cos 4 θ 2 d θ = 32 ∫ 0 π / 2 ( 1 − cos 4 θ ) d θ = 32 [ θ − sin 4 θ 4 ] 0 π / 2 = 32 ( π 2 − sin 2 π 4 ) = 16 π . \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} I = ∫ 0 π /2 16 sin 2 θ 16 − 16 sin 2 θ 4 cos θ d θ = ∫ 0 π /2 16 sin 2 θ ⋅ 4 cos θ ⋅ 4 cos θ d θ = 1 6 2 ∫ 0 π /2 sin 2 θ cos 2 θ d θ = 1 6 2 ∫ 0 π /2 ( 2 sin 2 θ ) 2 d θ = 4 1 6 2 ∫ 0 π /2 2 1 − cos 4 θ d θ = 32 ∫ 0 π /2 ( 1 − cos 4 θ ) d θ = 32 [ θ − 4 sin 4 θ ] 0 π /2 = 32 ( 2 π − 4 sin 2 π ) = 16 π . The calculation uses sin θ cos θ = sin 2 θ 2 \sin\theta\cos\theta=\frac{\sin 2\theta}{2} sin θ cos θ = 2 s i n 2 θ sin 2 2 θ = 1 − cos 4 θ 2 \sin^2 2\theta=\frac{1-\cos 4\theta}{2} sin 2 2 θ = 2 1 − c o s 4 θ
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Assume a > 0 a>0 a > 0 x = a tan θ x=a\tan\theta x = a tan θ − π 2 < θ < π 2 -\frac{\pi}{2}<\theta<\frac{\pi}{2} − 2 π < θ < 2 π d x = a sec 2 θ d θ dx=a\sec^2\theta\,d\theta d x = a sec 2 θ d θ sec θ > 0 \sec\theta>0 sec θ > 0
I = ∫ a sec 2 θ ( a 2 + a 2 tan 2 θ ) 3 / 2 d θ = 1 a 2 ∫ sec 2 θ ( 1 + tan 2 θ ) 3 / 2 d θ = 1 a 2 ∫ sec 2 θ sec 3 θ d θ = 1 a 2 ∫ 1 sec θ d θ = 1 a 2 ∫ cos θ d θ = 1 a 2 sin θ + C = x a 2 x 2 + a 2 + 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} I = ∫ ( a 2 + a 2 tan 2 θ ) 3/2 a sec 2 θ d θ = a 2 1 ∫ ( 1 + tan 2 θ ) 3/2 sec 2 θ d θ = a 2 1 ∫ sec 3 θ sec 2 θ d θ = a 2 1 ∫ sec θ 1 d θ = a 2 1 ∫ cos θ d θ = a 2 1 sin θ + C = a 2 x 2 + a 2 x + C . The identity sec 2 θ = tan 2 θ + 1 \sec^2\theta=\tan^2\theta+1 sec 2 θ = tan 2 θ + 1 ( 1 + tan 2 θ ) 3 / 2 = sec 3 θ (1+\tan^2\theta)^{3/2}=\sec^3\theta ( 1 + tan 2 θ ) 3/2 = sec 3 θ sec θ > 0 \sec\theta>0 sec θ > 0 cos θ = a / x 2 + a 2 \cos\theta=a/\sqrt{x^2+a^2} cos θ = a / x 2 + a 2 sin θ = tan θ cos θ = x / x 2 + a 2 \sin\theta=\tan\theta\cos\theta=x/\sqrt{x^2+a^2} sin θ = tan θ cos θ = x / x 2 + a 2 x < 0 x<0 x < 0
[RIGHT CLICK] [ASK AI] The same surds can be cleared by a hyperbolic identity. On the a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x = a sinh θ x=a\sinh\theta x = a sinh θ cosh 2 θ = 1 + sinh 2 θ \cosh^2\theta=1+\sinh^2\theta cosh 2 θ = 1 + sinh 2 θ sec 2 θ = 1 + tan 2 θ \sec^2\theta=1+\tan^2\theta sec 2 θ = 1 + tan 2 θ a cosh θ a\cosh\theta a cosh θ a sec θ a\sec\theta a sec θ
[ EXAMPLE ] Let x = 2 sinh θ x=2\sinh\theta x = 2 sinh θ d x = 2 cosh θ d θ dx=2\cosh\theta\,d\theta d x = 2 cosh θ d θ 4 + x 2 = 4 + 4 sinh 2 θ = 2 cosh θ \sqrt{4+x^2}=\sqrt{4+4\sinh^2\theta}=2\cosh\theta 4 + x 2 = 4 + 4 sinh 2 θ = 2 cosh θ cosh 2 θ − sinh 2 θ = 1 \cosh^2\theta-\sinh^2\theta=1 cosh 2 θ − sinh 2 θ = 1 cosh θ > 0 \cosh\theta>0 cosh θ > 0
I = ∫ 2 cosh θ 2 cosh θ d θ = θ + C = sinh − 1 ( x 2 ) + C . I=\int\frac{2\cosh\theta}{2\cosh\theta}\,d\theta=\theta+C=\sinh^{-1}\!\left(\frac{x}{2}\right)+C. I = ∫ 2 cosh θ 2 cosh θ d θ = θ + C = sinh − 1 ( 2 x ) + C . The trigonometric route on the same row, x = 2 tan θ x=2\tan\theta x = 2 tan θ ∫ sec θ d θ \int\sec\theta\,d\theta ∫ sec θ d θ
[RIGHT CLICK] [ASK AI] Part of the thread Accumulation and changes of variables
A change of variables redescribes the problem so the calculation is simpler, paying a scale factor such as a Jacobian.
Trigonometric and hyperbolic substitutions Not linked to the rest of the course yet.