U-Substitution Practice Problems

Questions

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∫cos⁡x esin⁡x dx\int \cos x\, e^{\sin x}\,dx

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∫2ax+bax2+bx+c dx(a≠0)\int \frac{2ax+b}{ax^2+bx+c}\,dx \quad(a\neq 0)

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∫cos⁡x (1+sin⁡2x) dx\int \cos x\,(1+\sin^2 x)\,dx

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∫cos⁡x  cos⁡5(sin⁡x) dx\int \cos x\;\cos^{5}(\sin x)\,dx

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∫dxxln⁡x(x>0, x≠1)\int \frac{dx}{x\ln x} \qquad(x>0,\ x\neq 1)

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∫x1+x2 dx\int \frac{x}{1+x^2}\,dx

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∫x 1+x2  dx\int \frac{x}{\sqrt{\,1+x^2\,}}\,dx

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∫ex1+ex dx\int \frac{e^{x}}{1+e^{x}}\,dx

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∫cos⁡x a+sin⁡x  dx(a∈R)\int \frac{\cos x}{\,a+\sin x\,}\,dx \quad(a\in\mathbb{R})

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∫kx+m(kx+m)2+c dx(k≠0)\int \frac{kx+m}{(kx+m)^2+c}\,dx \quad(k\neq 0)

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∫λ (2ax+b)ax2+bx+c dx(a≠0)\int \frac{\lambda\,(2ax+b)}{ax^2+bx+c}\,dx \quad(a\neq 0)

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∫(3x+1) e(3x+1)2 dx\int (3x+1)\,e^{(3x+1)^2}\,dx

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∫0π/2cos⁡x1+sin⁡x dx\int_{0}^{\pi/2}\frac{\cos x}{1+\sin x}\,dx

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∫ln⁡xx dx(x>0)\int \frac{\ln x}{x}\,dx \qquad(x>0)

 

Solutions

1) Let u=sin⁡xu=\sin x, du=cos⁡x dxdu=\cos x\,dx.

∫cos⁡x esin⁡x dx=∫eu du=eu+C=esin⁡x+C.\int \cos x\,e^{\sin x}\,dx=\int e^{u}\,du=e^{u}+C=e^{\sin x}+C.

2) Let u=ax2+bx+cu=ax^2+bx+c, du=(2ax+b) dxdu=(2ax+b)\,dx.

∫2ax+bax2+bx+c dx=∫duu=ln⁡∣u∣+C=ln⁡∣ax2+bx+c∣+C.\int \frac{2ax+b}{ax^2+bx+c}\,dx=\int \frac{du}{u}=\ln|u|+C=\ln|ax^2+bx+c|+C.

3) Let u=sin⁡xu=\sin x, du=cos⁡x dxdu=\cos x\,dx.

∫cos⁡x(1+sin⁡2x) dx=∫(1+u2) du=u+u33+C=sin⁡x+sin⁡3x3+C.\int \cos x(1+\sin^2 x)\,dx=\int (1+u^2)\,du=u+\tfrac{u^3}{3}+C =\sin x+\tfrac{\sin^3 x}{3}+C.

4) Let u=sin⁡xu=\sin x, du=cos⁡x dxdu=\cos x\,dx.

∫cos⁡x cos⁡5(sin⁡x) dx=∫cos⁡5u du=∫cos⁡4u  cos⁡u du.\int \cos x\,\cos^{5}(\sin x)\,dx=\int \cos^{5}u\,du =\int \cos^{4}u\;\cos u\,du.

Write cos⁡4u=(1−sin⁡2u)2\cos^{4}u=(1-\sin^{2}u)^2. Let w=sin⁡uw=\sin u, dw=cos⁡u dudw=\cos u\,du:

∫(1−w2)2 dw=w−23w3+15w5+C.\int (1-w^{2})^{2}\,dw =w-\tfrac{2}{3}w^{3}+\tfrac{1}{5}w^{5}+C.

Back-substitute w=sin⁡u, u=sin⁡xw=\sin u,\ u=\sin x:

sin⁡(sin⁡x)−23sin⁡3(sin⁡x)+15sin⁡5(sin⁡x)+C.\sin(\sin x)-\tfrac{2}{3}\sin^{3}(\sin x)+\tfrac{1}{5}\sin^{5}(\sin x)+C.

5) Let u=ln⁡xu=\ln x, du=dxxdu=\frac{dx}{x}.

∫dxxln⁡x=∫duu=ln⁡∣u∣+C=ln⁡∣ln⁡x∣+C.\int \frac{dx}{x\ln x}=\int \frac{du}{u}=\ln|u|+C=\ln|\ln x|+C.

6) Let u=1+x2u=1+x^2, du=2x dxdu=2x\,dx.

∫x1+x2 dx=12∫duu=12ln⁡(1+x2)+C.\int \frac{x}{1+x^2}\,dx=\tfrac12\int \frac{du}{u} =\tfrac12\ln(1+x^2)+C.

7) Let u=1+x2u=1+x^2, du=2x dxdu=2x\,dx.

∫x1+x2 dx=12∫u−1/2 du=1+x2+C.\int \frac{x}{\sqrt{1+x^2}}\,dx=\tfrac12\int u^{-1/2}\,du =\sqrt{1+x^2}+C.

8) Let u=1+exu=1+e^{x}, du=ex dxdu=e^{x}\,dx.

∫ex1+ex dx=∫duu=ln⁡(1+ex)+C.\int \frac{e^{x}}{1+e^{x}}\,dx=\int \frac{du}{u}=\ln(1+e^{x})+C.

9) Let u=a+sin⁡xu=a+\sin x, du=cos⁡x dxdu=\cos x\,dx.

∫cos⁡xa+sin⁡x dx=∫duu=ln⁡∣a+sin⁡x∣+C.\int \frac{\cos x}{a+\sin x}\,dx=\int \frac{du}{u}=\ln|a+\sin x|+C.

10) Let u=(kx+m)2+cu=(kx+m)^2+c. Then du=2(kx+m)k dx⇒(kx+m) dx=du2kdu=2(kx+m)k\,dx\Rightarrow (kx+m)\,dx=\dfrac{du}{2k}.

∫kx+m(kx+m)2+c dx=12k∫duu=12kln⁡ ⁣((kx+m)2+c)+C.\int \frac{kx+m}{(kx+m)^2+c}\,dx =\frac{1}{2k}\int \frac{du}{u} =\frac{1}{2k}\ln\!\big((kx+m)^2+c\big)+C.

11) Let u=ax2+bx+cu=ax^2+bx+c, du=(2ax+b) dxdu=(2ax+b)\,dx.

∫λ (2ax+b)ax2+bx+c dx=λ∫duu=λln⁡∣ax2+bx+c∣+C.\int \frac{\lambda\,(2ax+b)}{ax^2+bx+c}\,dx =\lambda\int \frac{du}{u} =\lambda\ln|ax^2+bx+c|+C.

12) Let u=(3x+1)2u=(3x+1)^2, du=6(3x+1) dx⇒(3x+1) dx=du6du=6(3x+1)\,dx\Rightarrow (3x+1)\,dx=\tfrac{du}{6}.

∫(3x+1)e(3x+1)2 dx=16∫eu du=16e(3x+1)2+C.\int (3x+1)e^{(3x+1)^2}\,dx=\tfrac16\int e^{u}\,du =\tfrac16 e^{(3x+1)^2}+C.

13) Let u=1+sin⁡xu=1+\sin x, du=cos⁡x dxdu=\cos x\,dx.

∫0π/2cos⁡x1+sin⁡x dx=∫u=1u=2duu=ln⁡2.\int_{0}^{\pi/2}\frac{\cos x}{1+\sin x}\,dx =\int_{u=1}^{u=2}\frac{du}{u} =\ln 2.

14) Let u=ln⁡xu=\ln x, du=dxxdu=\frac{dx}{x}.

∫ln⁡xx dx=∫u du=12(ln⁡x)2+C.\int \frac{\ln x}{x}\,dx=\int u\,du=\tfrac12(\ln x)^2+C.

 

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