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[1251]

Examples from Reduction formulae

All examples from Reduction formulae.

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∫x3ex dx\int x^3 e^x \, dx

The recurrence with n=3n=3, then n=2n=2, then n=1n=1 and I0=∫ex dxI_0=\int e^x \, dx, gives

I3=x3ex−3I2=x3ex−3(x2ex−2I1)=x3ex−3x2ex+6I1=x3ex−3x2ex+6(xex−I0)=x3ex−3x2ex+6xex−6∫ex dx=x3ex−3x2ex+6xex−6ex+C.\begin{aligned} I_3&=x^3 e^x-3 I_2 \\ &=x^3 e^x-3\bigl(x^2 e^x-2 I_1\bigr) \\ &=x^3 e^x-3x^2 e^x+6 I_1 \\ &=x^3 e^x-3x^2 e^x+6\bigl(x e^x-I_0\bigr) \\ &=x^3 e^x-3x^2 e^x+6x e^x-6\int e^x \, dx \\ &=x^3 e^x-3x^2 e^x+6x e^x-6e^x+C. \end{aligned}

∫0π/2sin⁡5x dx\int_0^{\pi/2}\sin^5 x \, dx

I5=45I3=45⋅23I1=815∫0π/2sin⁡x dx=815[−cos⁡x]0π/2=815.I_5=\frac{4}{5} I_3=\frac{4}{5}\cdot\frac{2}{3} I_1=\frac{8}{15}\int_0^{\pi/2}\sin x \, dx=\frac{8}{15}\bigl[-\cos x\bigr]_0^{\pi/2}=\frac{8}{15}.

∫0π/2cos⁡7x dx\int_0^{\pi/2}\cos^7 x \, dx

J7=67J5=67⋅45J3=2435⋅23J1=1635∫0π/2cos⁡x dx=1635[sin⁡x]0π/2=1635.J_7=\frac{6}{7} J_5=\frac{6}{7}\cdot\frac{4}{5} J_3=\frac{24}{35}\cdot\frac{2}{3} J_1=\frac{16}{35}\int_0^{\pi/2}\cos x \, dx=\frac{16}{35}\bigl[\sin x\bigr]_0^{\pi/2}=\frac{16}{35}.

Secant and tangent recurrences on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

The base value I2=∫0π/4sec⁡2x dx=[tan⁡x]0π/4=1I_2=\int_0^{\pi/4}\sec^2 x \, dx=\bigl[\tan x\bigr]_0^{\pi/4}=1 gives

∫0π/4sec⁡4x dx=I4=13(2(4−2)/2+2I2)=13(2+2)=43.\int_0^{\pi/4}\sec^4 x \, dx=I_4=\frac{1}{3}\bigl(2^{(4-2)/2}+2I_2\bigr)=\frac{1}{3}(2+2)=\frac{4}{3}.

The base value J2=∫0π/4tan⁡2x dx=∫0π/4(sec⁡2x−1) dx=[tan⁡x−x]0π/4=1−π4J_2=\int_0^{\pi/4}\tan^2 x \, dx=\int_0^{\pi/4}(\sec^2 x-1)\, dx=\bigl[\tan x-x\bigr]_0^{\pi/4}=1-\frac{\pi}{4} gives

∫0π/4tan⁡4x dx=J4=13−J2=13−(1−π4)=π4−23.\int_0^{\pi/4}\tan^4 x \, dx=J_4=\frac{1}{3}-J_2=\frac{1}{3}-\bigl(1-\frac{\pi}{4}\bigr)=\frac{\pi}{4}-\frac{2}{3}.

For mixed powers, K1=∫0π/4tan⁡θsec⁡θ dθ=[sec⁡θ]0π/4=2−1K_1=\int_0^{\pi/4}\tan\theta\sec\theta \, d\theta=\bigl[\sec\theta\bigr]_0^{\pi/4}=\sqrt{2}-1, so

∫0π/4tan⁡3θsec⁡θ dθ=K3=13(2−2K1)=13(2−2(2−1))=2−23.\int_0^{\pi/4}\tan^3\theta\sec\theta \, d\theta=K_3=\frac{1}{3}\bigl(\sqrt{2}-2K_1\bigr)=\frac{1}{3}\bigl(\sqrt{2}-2(\sqrt{2}-1)\bigr)=\frac{2-\sqrt{2}}{3}.

∫0π/2cos⁡5xsin⁡4x dx\int_0^{\pi/2}\cos^5 x\sin^4 x \, dx

This is the definite version of the odd-power evaluation of ∫cos⁡5xsin⁡4x dx\int\cos^5 x\sin^4 x \, dx. Reduce the cosine power twice, then the sine power twice and use I1,0=1I_{1,0}=1:

I5,4=49I3,4=49⋅27I1,4=863I1,4=863⋅35I1,2=8105I1,2=8105⋅13I1,0=8315.\begin{aligned} I_{5,4}&=\frac{4}{9} I_{3,4}=\frac{4}{9}\cdot\frac{2}{7} I_{1,4}=\frac{8}{63} I_{1,4} \\ &=\frac{8}{63}\cdot\frac{3}{5} I_{1,2}=\frac{8}{105} I_{1,2} \\ &=\frac{8}{105}\cdot\frac{1}{3} I_{1,0}=\frac{8}{315}. \end{aligned}

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