08. 적분의 계산Techniques of Integration

Solution.

(1) Since $\dfrac{d}{dx}\left(xe^{x^3}\right) = e^{x^3}+x\cdot3x^2e^{x^3} = (1+3x^3)e^{x^3}$, $$\int (1+3x^3)e^{x^3}\,dx = xe^{x^3}+C.$$ (Check: $\dfrac{d}{dx}\left(xe^{x^3}+C\right) = (1+3x^3)e^{x^3}$.)

(2) By parts with $u=x^2, dv=e^x\,dx$, so $du=2x\,dx, v=e^x$, $$\int x^2e^x\,dx = x^2e^x-2\int xe^x\,dx,$$ and by parts again, $\int xe^x\,dx = xe^x-e^x+C$. Hence $$\int x^2e^x\,dx = x^2e^x-2(xe^x-e^x)+C = (x^2-2x+2)e^x+C.$$ (Check: $\dfrac{d}{dx}\left[(x^2-2x+2)e^x\right] = (2x-2)e^x+(x^2-2x+2)e^x = x^2e^x$.)

(3) By parts with $u=\cos^{-1}x, dv=dx$, so $du=-\dfrac1{\sqrt{1-x^2}}\,dx, v=x$, and using $\displaystyle\int \frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}+C$, $$\int \cos^{-1}x\,dx = x\cos^{-1}x+\int \frac{x}{\sqrt{1-x^2}}\,dx = x\cos^{-1}x-\sqrt{1-x^2}+C.$$ (Check: $\dfrac{d}{dx}\left[x\cos^{-1}x-\sqrt{1-x^2}\right] = \cos^{-1}x-\dfrac{x}{\sqrt{1-x^2}}+\dfrac{x}{\sqrt{1-x^2}} = \cos^{-1}x$.)

(4) By parts with $u=\sinh^{-1}x, dv=dx$, so $du=\dfrac1{\sqrt{1+x^2}}\,dx, v=x$, $$\int \sinh^{-1}x\,dx = x\sinh^{-1}x-\int \frac{x}{\sqrt{1+x^2}}\,dx = x\sinh^{-1}x-\sqrt{1+x^2}+C.$$ (Check: $\dfrac{d}{dx}\left[x\sinh^{-1}x-\sqrt{1+x^2}\right] = \sinh^{-1}x+\dfrac{x}{\sqrt{1+x^2}}-\dfrac{x}{\sqrt{1+x^2}} = \sinh^{-1}x$.)

Proof.

(1) Write $\cos^n x = \cos^{n-1}x\cdot\cos x$ and integrate by parts with $u=\cos^{n-1}x, dv=\cos x\,dx$, so $du=-(n-1)\cos^{n-2}x\sin x\,dx, v=\sin x$. Then $$\int \cos^n x\,dx = \cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x\sin^2x\,dx.$$ Substituting $\sin^2x=1-\cos^2x$, $$\int \cos^n x\,dx = \cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x\,dx-(n-1)\int \cos^n x\,dx,$$ so collecting the $\int \cos^n x\,dx$ terms on the left gives $$n\int \cos^n x\,dx = \cos^{n-1}x\sin x+(n-1)\int \cos^{n-2}x\,dx,$$ that is, $$\int \cos^n x\,dx = \frac{\cos^{n-1}x\sin x}{n}+\frac{n-1}{n}\int \cos^{n-2}x\,dx.$$

(2) Write $\sin^nx\cos^mx = \sin^{n-1}x\cdot(\sin x\cos^mx)$ and integrate by parts with $u=\sin^{n-1}x, dv=\sin x\cos^mx\,dx$. Then $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=\displaystyle\int \sin x\cos^mx\,dx = -\frac{\cos^{m+1}x}{m+1}$, so $$I:=\int \sin^nx\cos^mx\,dx = -\frac{\sin^{n-1}x\cos^{m+1}x}{m+1}+\frac{n-1}{m+1}\int \sin^{n-2}x\cos^{m+2}x\,dx.$$ Since $\cos^{m+2}x=\cos^mx(1-\sin^2x)$, $$\int \sin^{n-2}x\cos^{m+2}x\,dx = \underbrace{\int \sin^{n-2}x\cos^mx\,dx}_{=:J}-I,$$ and substituting this in gives $$I = -\frac{\sin^{n-1}x\cos^{m+1}x}{m+1}+\frac{n-1}{m+1}(J-I).$$ Collecting the $I$ terms on the left, $$I\left(1+\frac{n-1}{m+1}\right) = -\frac{\sin^{n-1}x\cos^{m+1}x}{m+1}+\frac{n-1}{m+1}J,$$ and since $1+\dfrac{n-1}{m+1}=\dfrac{m+n}{m+1}$, multiplying both sides by $\dfrac{m+1}{m+n}$ gives $$I = -\frac{\sin^{n-1}x\cos^{m+1}x}{m+n}+\frac{n-1}{m+n}J = -\frac{\sin^{n-1}x\cos^{m+1}x}{m+n}+\frac{n-1}{m+n}\int \sin^{n-2}x\cos^mx\,dx.$$

(3) Integrate by parts with $u=(\ln x)^n, dv=dx$, so $du=n(\ln x)^{n-1}\cdot\dfrac1x\,dx, v=x$. Then $$\int (\ln x)^n\,dx = x(\ln x)^n-\int x\cdot n(\ln x)^{n-1}\cdot\frac1x\,dx = x(\ln x)^n-n\int (\ln x)^{n-1}\,dx.$$

(4) Let $F(x):=\displaystyle\int_x^b f(t)\,dt$, so $F'(x)=-f(x)$. Integrate by parts with $u=F(x), dv=dx$, choosing $v=x-a$ (any constant works since $dv=dx$). Then $$\int_a^b F(x)\,dx = \Big[(x-a)F(x)\Big]_a^b - \int_a^b (x-a)F'(x)\,dx = \Big[(x-a)F(x)\Big]_a^b + \int_a^b (x-a)f(x)\,dx.$$ At $x=b$, $(b-a)F(b)=(b-a)\displaystyle\int_b^bf(t)\,dt=0$, and at $x=a$, $(a-a)F(a)=0$, so the boundary term vanishes. Hence $$\int_a^b\left(\int_x^bf(t)\,dt\right)dx = \int_a^b(x-a)f(x)\,dx.$$

Solution.

(1) By the product-to-sum formula $\sin A\cos B=\dfrac12\left[\sin(A+B)+\sin(A-B)\right]$, $$\sin3x\cos5x = \frac12\left[\sin8x+\sin(-2x)\right] = \frac12(\sin8x-\sin2x),$$ so $$\int \sin3x\cos5x\,dx = \frac12\int(\sin8x-\sin2x)\,dx = \frac12\left(-\frac{\cos8x}8+\frac{\cos2x}2\right)+C = \frac{\cos2x}4-\frac{\cos8x}{16}+C.$$ (Check: $\dfrac{d}{dx}\left[\dfrac{\cos2x}4-\dfrac{\cos8x}{16}\right] = -\dfrac{\sin2x}2+\dfrac{\sin8x}2 = \dfrac12(\sin8x-\sin2x) = \sin3x\cos5x$.)

(2) By the product-to-sum formula $\cos A\cos B=\dfrac12\left[\cos(A-B)+\cos(A+B)\right]$, $$\cos x\cos7x = \frac12(\cos6x+\cos8x),$$ so $$\int_{-\pi/2}^{\pi/2}\cos x\cos7x\,dx = \left[\frac{\sin6x}{12}+\frac{\sin8x}{16}\right]_{-\pi/2}^{\pi/2}.$$ At $x=\pi/2$, $\sin(3\pi)=0$ and $\sin(4\pi)=0$; at $x=-\pi/2$, $\sin(-3\pi)=0$ and $\sin(-4\pi)=0$ as well, so both endpoint values are $0$. Hence $$\int_{-\pi/2}^{\pi/2}\cos x\cos7x\,dx = 0-0 = 0.$$

Solution. By $\sin A\sin B=\dfrac12\left[\cos(A-B)-\cos(A+B)\right]$, $$\sin\theta\sin3\theta = \frac12(\cos2\theta-\cos4\theta),$$ so $$\sin\theta\sin2\theta\sin3\theta = \sin2\theta\cdot\frac12(\cos2\theta-\cos4\theta) = \frac12\sin2\theta\cos2\theta-\frac12\sin2\theta\cos4\theta.$$ Now $\sin2\theta\cos2\theta=\dfrac12\sin4\theta$, and $\sin2\theta\cos4\theta=\dfrac12\left[\sin6\theta+\sin(-2\theta)\right]=\dfrac12(\sin6\theta-\sin2\theta)$, so $$\sin\theta\sin2\theta\sin3\theta = \frac14\sin4\theta-\frac14(\sin6\theta-\sin2\theta) = \frac14\sin2\theta+\frac14\sin4\theta-\frac14\sin6\theta.$$ Integrating term by term, $$\int \sin\theta\sin2\theta\sin3\theta\,d\theta = -\frac{\cos2\theta}8-\frac{\cos4\theta}{16}+\frac{\cos6\theta}{24}+C.$$ (Check: $\dfrac{d}{d\theta}\left[-\dfrac{\cos2\theta}8-\dfrac{\cos4\theta}{16}+\dfrac{\cos6\theta}{24}\right] = \dfrac{\sin2\theta}4+\dfrac{\sin4\theta}4-\dfrac{\sin6\theta}4 = \sin\theta\sin2\theta\sin3\theta$.)

Solution.

(1) Substituting $x=2\tan\theta$ gives $dx=2\sec^2\theta\,d\theta$, $\sqrt{4+x^2}=2\sec\theta$, so $$\int \frac{dx}{\sqrt{4+x^2}} = \int \frac{2\sec^2\theta\,d\theta}{2\sec\theta} = \int \sec\theta\,d\theta = \ln|\sec\theta+\tan\theta|+C.$$ Since $\tan\theta=\dfrac x2$ and $\sec\theta=\dfrac{\sqrt{4+x^2}}2$, $$\int \frac{dx}{\sqrt{4+x^2}} = \ln\left(\frac{x+\sqrt{4+x^2}}2\right)+C = \ln\left(x+\sqrt{4+x^2}\right)+C = \sinh^{-1}\frac x2+C.$$ (Check: $\dfrac{d}{dx}\sinh^{-1}\dfrac x2 = \dfrac{1/2}{\sqrt{1+x^2/4}} = \dfrac1{\sqrt{4+x^2}}$.)

(2) Substituting $x=3\sin\theta$ gives $dx=3\cos\theta\,d\theta$, $\sqrt{9-x^2}=3\cos\theta$, so $$\int \frac{x^2\,dx}{\sqrt{9-x^2}} = \int \frac{9\sin^2\theta\cdot3\cos\theta}{3\cos\theta}\,d\theta = 9\int \sin^2\theta\,d\theta = \frac92\int(1-\cos2\theta)\,d\theta = \frac92\theta-\frac94\sin2\theta+C.$$ Since $\sin2\theta=2\sin\theta\cos\theta$ and $\theta=\sin^{-1}\dfrac x3, \sin\theta=\dfrac x3, \cos\theta=\dfrac{\sqrt{9-x^2}}3$, $$\frac94\sin2\theta = \frac94\cdot2\cdot\frac x3\cdot\frac{\sqrt{9-x^2}}3 = \frac{x\sqrt{9-x^2}}2,$$ and therefore $$\int \frac{x^2\,dx}{\sqrt{9-x^2}} = \frac92\sin^{-1}\frac x3-\frac x2\sqrt{9-x^2}+C.$$ (Check: $\dfrac{d}{dx}\left[\dfrac92\sin^{-1}\dfrac x3-\dfrac x2\sqrt{9-x^2}\right] = \dfrac{9/2}{\sqrt{9-x^2}}+\dfrac{x^2-9/2}{\sqrt{9-x^2}} = \dfrac{x^2}{\sqrt{9-x^2}}$.)

(3) Since $25x^2-4=4\left(\dfrac{25x^2}4-1\right)$, substituting $u=5x$ gives $du=5\,dx$, so $$\int \frac{dx}{\sqrt{25x^2-4}} = \int \frac{du/5}{\sqrt{u^2-4}} = \frac15\int \frac{du}{\sqrt{u^2-4}} = \frac15\cosh^{-1}\frac u2+C = \frac15\cosh^{-1}\frac{5x}2+C.$$ Since $\cosh^{-1}\dfrac{5x}2=\ln\left(\dfrac{5x+\sqrt{25x^2-4}}2\right)$, this can also be written as $$\int \frac{dx}{\sqrt{25x^2-4}} = \frac15\ln\left|5x+\sqrt{25x^2-4}\right|+C.$$ (Check: $\dfrac{d}{dx}\left[\dfrac15\ln\left|5x+\sqrt{25x^2-4}\right|\right] = \dfrac15\cdot\dfrac{5+\frac{25x}{\sqrt{25x^2-4}}}{5x+\sqrt{25x^2-4}} = \dfrac15\cdot\dfrac{5\left(\sqrt{25x^2-4}+5x\right)}{\sqrt{25x^2-4}\left(5x+\sqrt{25x^2-4}\right)} = \dfrac1{\sqrt{25x^2-4}}$.)

Solution.

(1) Substituting $u=\ln x$ gives $du=\dfrac1x\,dx$, so $$\int \frac{\sqrt{1-(\ln x)^2}}{x\ln x}\,dx = \int \frac{\sqrt{1-u^2}}u\,du.$$ Substituting again $u=\sin\phi$ gives $du=\cos\phi\,d\phi, \sqrt{1-u^2}=\cos\phi$, so $$\int \frac{\sqrt{1-u^2}}u\,du = \int \frac{\cos^2\phi}{\sin\phi}\,d\phi = \int\left(\csc\phi-\sin\phi\right)d\phi = \ln|\csc\phi-\cot\phi|+\cos\phi+C.$$ Since $\sin\phi=u, \cos\phi=\sqrt{1-u^2}$, we have $\csc\phi-\cot\phi=\dfrac{1-\sqrt{1-u^2}}u=\dfrac u{1+\sqrt{1-u^2}}$, and simplifying inside the logarithm, $$\int \frac{\sqrt{1-u^2}}u\,du = \sqrt{1-u^2}-\ln\left|\frac{1+\sqrt{1-u^2}}u\right|+C.$$ Substituting $u=\ln x$, $$\int \frac{\sqrt{1-(\ln x)^2}}{x\ln x}\,dx = \sqrt{1-(\ln x)^2}-\ln\left|\frac{1+\sqrt{1-(\ln x)^2}}{\ln x}\right|+C.$$ (Check: $\dfrac{d}{du}\left[\sqrt{1-u^2}-\ln\left(1+\sqrt{1-u^2}\right)+\ln|u|\right] = \dfrac{-u}{\sqrt{1-u^2}}+\dfrac u{\sqrt{1-u^2}\left(1+\sqrt{1-u^2}\right)}+\dfrac1u = \dfrac{-u}{1+\sqrt{1-u^2}}+\dfrac1u = \dfrac{\sqrt{1-u^2}}u$, and since $u=\ln x$, the chain rule gives $\dfrac{d}{dx}[\cdots] = \dfrac{\sqrt{1-(\ln x)^2}}{\ln x}\cdot\dfrac1x = \dfrac{\sqrt{1-(\ln x)^2}}{x\ln x}$.)

(2) Substituting $x=4\sin^2\theta$ $(0\lt\theta\lt\pi/2)$ gives $dx=8\sin\theta\cos\theta\,d\theta$ and $4-x=4\cos^2\theta$, so $$\sqrt{\frac{4-x}x} = \sqrt{\frac{4\cos^2\theta}{4\sin^2\theta}} = \frac{\cos\theta}{\sin\theta}.$$ Hence $$\int \sqrt{\frac{4-x}x}\,dx = \int \frac{\cos\theta}{\sin\theta}\cdot8\sin\theta\cos\theta\,d\theta = 8\int \cos^2\theta\,d\theta = 4\theta+4\sin\theta\cos\theta+C.$$ Since $\sin\theta=\dfrac{\sqrt x}2, \cos\theta=\dfrac{\sqrt{4-x}}2, \theta=\sin^{-1}\dfrac{\sqrt x}2$, we have $4\sin\theta\cos\theta=\sqrt{x(4-x)}$, so $$\int \sqrt{\frac{4-x}x}\,dx = 4\sin^{-1}\frac{\sqrt x}2+\sqrt{x(4-x)}+C.$$ (Check: $\dfrac{d}{dx}\left[4\sin^{-1}\dfrac{\sqrt x}2\right] = \dfrac2{\sqrt{x(4-x)}}$ and $\dfrac{d}{dx}\sqrt{x(4-x)} = \dfrac{2-x}{\sqrt{x(4-x)}}$, so the sum of the two derivatives is $\dfrac{4-x}{\sqrt{x(4-x)}} = \sqrt{\dfrac{4-x}x}$.)

(3) Substituting $u=x^{3/2}$ gives $du=\dfrac32\sqrt x\,dx$, i.e. $\sqrt x\,dx=\dfrac23\,du$, and $u^2=x^3$, so $$\int \sqrt{\frac x{1-x^3}}\,dx = \int \frac{\sqrt x\,dx}{\sqrt{1-x^3}} = \int \frac{(2/3)\,du}{\sqrt{1-u^2}} = \frac23\sin^{-1}u+C = \frac23\sin^{-1}\left(x^{3/2}\right)+C.$$ (Check: $\dfrac{d}{dx}\left[\dfrac23\sin^{-1}\left(x^{3/2}\right)\right] = \dfrac23\cdot\dfrac{(3/2)x^{1/2}}{\sqrt{1-x^3}} = \dfrac{\sqrt x}{\sqrt{1-x^3}} = \sqrt{\dfrac x{1-x^3}}$.)

(4) Completing the square, $x^2-2x+5=(x-1)^2+4$, and substituting $w=x-1$, $$\int \frac1{\sqrt{x^2-2x+5}}\,dx = \int \frac{dw}{\sqrt{w^2+4}} = \sinh^{-1}\frac w2+C = \sinh^{-1}\frac{x-1}2+C.$$ This can also be written as $\ln\left|w+\sqrt{w^2+4}\right|+C = \ln\left|x-1+\sqrt{x^2-2x+5}\right|+C$. (Check: $\dfrac{d}{dx}\ln\left|x-1+\sqrt{x^2-2x+5}\right| = \dfrac{1+\frac{x-1}{\sqrt{x^2-2x+5}}}{x-1+\sqrt{x^2-2x+5}} = \dfrac{\sqrt{x^2-2x+5}+x-1}{\sqrt{x^2-2x+5}\left(x-1+\sqrt{x^2-2x+5}\right)} = \dfrac1{\sqrt{x^2-2x+5}}$.)

Solution. We check that all three methods give the same result.

(Method 1: substitution) Substituting $u=1-x^2$ gives $du=-2x\,dx$, $x^2=1-u$, so $$x^3\sqrt{1-x^2}\,dx = x^2\sqrt{1-x^2}\cdot x\,dx = (1-u)\sqrt u\left(-\frac{du}2\right) = \left(-\frac12u^{1/2}+\frac12u^{3/2}\right)du,$$ and hence $$\int x^3\sqrt{1-x^2}\,dx = -\frac13u^{3/2}+\frac15u^{5/2}+C = \frac{(1-x^2)^{5/2}}5-\frac{(1-x^2)^{3/2}}3+C.$$

(Method 2: integration by parts) Let $u=x^2, dv=x\sqrt{1-x^2}\,dx$, so $du=2x\,dx$ and $v=\displaystyle\int x\sqrt{1-x^2}\,dx=-\dfrac13(1-x^2)^{3/2}$. Then $$\int x^3\sqrt{1-x^2}\,dx = -\frac{x^2}3(1-x^2)^{3/2}+\frac23\int x(1-x^2)^{3/2}\,dx.$$ Since $\displaystyle\int x(1-x^2)^{3/2}\,dx=-\dfrac15(1-x^2)^{5/2}+C$, $$\int x^3\sqrt{1-x^2}\,dx = -\frac{x^2}3(1-x^2)^{3/2}-\frac2{15}(1-x^2)^{5/2}+C.$$ Substituting $x^2=1-(1-x^2)$ and simplifying shows this is exactly the same expression as the result of Method 1, $\dfrac{(1-x^2)^{5/2}}5-\dfrac{(1-x^2)^{3/2}}3+C$.

(Method 3: trigonometric substitution) Substituting $x=\sin\theta$ gives $dx=\cos\theta\,d\theta, \sqrt{1-x^2}=\cos\theta$, so $$\int x^3\sqrt{1-x^2}\,dx = \int \sin^3\theta\cos^2\theta\,d\theta = \int \sin\theta\left(\cos^2\theta-\cos^4\theta\right)d\theta.$$ Substituting $c=\cos\theta$ gives $dc=-\sin\theta\,d\theta$, so $$\int \sin\theta\left(\cos^2\theta-\cos^4\theta\right)d\theta = -\int\left(c^2-c^4\right)dc = -\frac{c^3}3+\frac{c^5}5+C = \frac{\cos^5\theta}5-\frac{\cos^3\theta}3+C,$$ and since $\cos\theta=\sqrt{1-x^2}$, $$\int x^3\sqrt{1-x^2}\,dx = \frac{(1-x^2)^{5/2}}5-\frac{(1-x^2)^{3/2}}3+C,$$ the same result as Methods 1 and 2.

(Check: $\dfrac{d}{dx}\left[\dfrac{(1-x^2)^{5/2}}5-\dfrac{(1-x^2)^{3/2}}3\right] = -x(1-x^2)^{3/2}+x(1-x^2)^{1/2} = x(1-x^2)^{1/2}\left[1-(1-x^2)\right] = x^3\sqrt{1-x^2}$.)